i have this string:
12 4 the quick 99 -1 fox dog \
what i want in my program:
myArray[] = {12, 4, 99, -1};
how i do a multiple number scanning?
See my answer to your other question here. It's a relatively simple matter to replace the strtok section to recognize non-numeric words and neither increment the count (in the first pass) nor load them into the array (in the second pass).
The code has changed as follows:
Using an input file of:
12 3 45 6 7 8
3 5 6 7
7 0 -1 4 5
12 4 the quick 99 -1 fox dog \
it produces output along the lines of:
0x8e42170, size = 6:
12 3 45 6 7 8
0x8e421d0, size = 4:
3 5 6 7
0x8e421e0, size = 5:
7 0 -1 4 5
0x8e42278, size = 4:
12 4 99 -1
Here's the code that produced that output:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <errno.h>
// This is the linked list of integer arrays.
typedef struct _tIntArray {
int size;
int *array;
struct _tIntArray *next;
} tIntArray;
static tIntArray *first = NULL;
static tIntArray *last = NULL;
// Check that argument is numeric, optional minus sign followed by
// zero or more digits (you may want one or more).
static int isAllNumeric (char *word) {
char *s = word;
if (*s == '-')
s++;
for (; *s != '\0'; s++)
if ((*s < '0') || (*s > '9'))
return 0;
return 1;
}
// Add a line of integers as a node.
static int addNode (char *str) {
tIntArray *curr; // pointers for new integer array.
char *word; // word within string.
char *tmpStr; // temp copy of buffer.
int fldCnt; // field count for line.
int i;
// Count number of fields.
if ((tmpStr = strdup (str)) == NULL) {
printf ("Cannot allocate duplicate string (%d).\n", errno);
return 1;
}
fldCnt = 0;
for (word = strtok (tmpStr, " "); word; word = strtok (NULL, " "))
if (isAllNumeric (word))
fldCnt++;
free (tmpStr);
// Create new linked list node.
if ((curr = malloc (sizeof (tIntArray))) == NULL) {
printf ("Cannot allocate integer array node (%d).\n", errno);
return 1;
}
curr->size = fldCnt;
if ((curr->array = malloc (fldCnt * sizeof (int))) == NULL) {
printf ("Cannot allocate integer array (%d).\n", errno);
free (curr);
return 1;
}
curr->next = NULL;
for (i = 0, word = strtok (str, " "); word; word = strtok (NULL, " "))
if (isAllNumeric (word))
curr->array[i++] = atoi (word);
if (last == NULL)
first = last = curr;
else {
last->next = curr;
last = curr;
}
return 0;
}
int main(void) {
int lineSz; // current line size.
char *buff; // buffer to hold line.
FILE *fin; // input file handle.
long offset; // offset for re-allocating line buffer.
tIntArray *curr; // pointers for new integer array.
int i;
// Open file.
if ((fin = fopen ("qq.in", "r")) == NULL) {
printf ("Cannot open qq.in, errno = %d\n", errno);
return 1;
}
// Allocate initial line.
lineSz = 2;
if ((buff = malloc (lineSz+1)) == NULL) {
printf ("Cannot allocate initial memory, errno = %d.\n", errno);
return 1;
}
// Loop forever.
while (1) {
// Save offset in case we need to re-read.
offset = ftell (fin);
// Get line, exit if end of file.
if (fgets (buff, lineSz, fin) == NULL)
break;
// If no newline, assume buffer wasn't big enough.
if (buff[strlen(buff)-1] != '\n') {
// Get bigger buffer and seek back to line start and retry.
free (buff);
lineSz += 3;
if ((buff = malloc (lineSz+1)) == NULL) {
printf ("Cannot allocate extra memory, errno = %d.\n", errno);
return 1;
}
if (fseek (fin, offset, SEEK_SET) != 0) {
printf ("Cannot seek, errno = %d.\n", errno);
return 1;
}
continue;
}
// Remove newline and process.
buff[strlen(buff)-1] = '\0';
if (addNode (buff) != 0)
return 1;
}
// Dump table for debugging.
for (curr = first; curr != NULL; curr = curr->next) {
printf ("%p, size = %d:\n ", curr, curr->size);
for (i = 0; i < curr->size; i++)
printf (" %d", curr->array[i]);
printf ("\n");
}
// Free resources and exit.
free (buff);
fclose (fin);
return 0;
}
Related
I get the error "Cannot access memory at address 0x100403055" when I try and set a memory value to 0x00 when stopped in the debugger.
Is there a special switch I need to set to enable the set operation?
Here is my complete C code file "main.c"
#include <stdio.h>
#include <string.h>
/*
separator - consume all non-token characters until next token. This includes:
comments: '#'
nesting: '{'
unnesting: '}'
whitespace: ' ','\t','\n'
*nest is changed according to nesting/unnesting processed
*/
static void separator(int *nest, char **tokens) {
char c, *s;
s = *tokens;
while ((c = *s)) {
/* #->eol = comment */
if (c == '#') {
s++;
while ((c = *s)) {
s++;
if (c == '\n')
break;
}
continue;
}
if (c == '{') {
(*nest)++;
s++;
continue;
}
if (c == '}') {
(*nest)--;
s++;
continue;
}
if (c == ' ' || c == '\n' || c == '\t') {
s++;
continue;
}
break;
}
*tokens = s;
}
/*
token - capture all characters until next separator, then consume separator,
return captured token, leave **tokens pointing to next token.
*/
static char *token(int *nest, char **tokens) {
char c, *s, *t;
char terminator = '\0';
s = t = *tokens;
while ((c = *s)) {
if (c == '#'
|| c == ' ' || c == '\t' || c == '\n' || c == '{' || c == '}')
break;
s++;
}
*tokens = s;
separator(nest, tokens);
/* Breakpoint here to examine and manipulate memory */
*s = '\0';
return t;
}
struct test_case {
char *input;
int nest;
char *expected_output;
};
int main() {
int nest = 0;
int TESTSEP = 0;
if (TESTSEP>0) {
char *tokens = "# this is a comment\n{nesting {example} unnesting}\n \t end";
separator(&nest, &tokens);
printf("nest: %d\n", nest);
printf("tokens: %s\n", tokens);
return 0;
} else {
struct test_case test_cases[] = {
{"hello world", 0, "hello"},
{"hello#world", 0, "hello"},
{"hello{world}", 0, "hello"},
{"hello world", 0, "hello"},
{"hello\tworld", 0, "hello"},
{"hello\nworld", 0, "hello"},
};
for (int i = 0; i < sizeof(test_cases) / sizeof(test_cases[0]); i++) {
struct test_case test_case = test_cases[i];
char *tokens = test_case.input;
char *output = token(&test_case.nest, &tokens);
if (strcmp(output, test_case.expected_output) != 0) {
printf("Test case %d failed: expected %s, got %s\n", i, test_case.expected_output, output);
}
}
return 0;
}
}
In the token function there is a comment line where I place a breakpoint and drop into the gdb debugger. The code is supposed to write a '\0' at the address of the pointer *s to truncate the string.
When I'm in the debugger and I examine the 's' variable I get the following:
(gdb) x s
0x100403055: 0x726f7720
When I try and set the variable I get:
(gdb) [![set *0x0000000100403055 = 0x726f7700][1]][1]
Cannot access memory at address 0x100403055
I'm using the CLION IDE and am a novice. I'm not sure if its an IDE problem, a user problem or some external memory protection mechanism that is preventing this.
Does anyone know how to make this work?
Here is a screenshot of the IDE:
When I run the code (without the debugger) I get this output:
./explore.exe
Test case 0 failed: expected hello, got hello world
Test case 1 failed: expected hello, got hello#world
Test case 2 failed: expected hello, got hello{world}
Test case 3 failed: expected hello, got hello world
Test case 4 failed: expected hello, got hello world
Test case 5 failed: expected hello, got hello world
Process finished with exit code 0
I this case I believe I was passing in a pointer to memory in the read only space. The struct test_case is built into the code and is read only. So that when I pass that into the token function it was trying to write to read only.
Here is the code that seems to work.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
/*
separator - consume all non-token characters until next token.
This includes:
comments: '#' ... '\n'
nesting: '{'
unnesting: '}'
whitespace: ' ','\t','\n'
*nest is changed according to nesting/unnesting processed
*/
static void separator(int *nest, char **tokens) {
char c, *s;
s = *tokens;
while ((c = *s)) {
/* #->eol = comment */
if (c == '#') {
s++;
while ((c = *s)) {
s++;
if (c == '\n')
break;
}
continue;
}
if (c == '{') {
(*nest)++;
s++;
continue;
}
if (c == '}') {
(*nest)--;
s++;
continue;
}
if (c == ' ' || c == '\n' || c == '\t') {
s++;
continue;
}
break;
}
*tokens = s;
}
/*
token - capture all characters until next separator, then consume
separator,
return captured token, leave **tokens pointing to next token.
*/
static char *token(int *nest, char **tokens) {
char c, *s, *t;
char terminator = '\0';
s = t = *tokens;
while ((c = *s)) {
if (c == '#'
|| c == ' ' || c == '\t' || c == '\n' || c == '{' || c == '}')
break;
s++;
}
*tokens = s;
separator(nest, tokens);
*s = '\0';
return t;
}
struct test_case {
char *input;
int nest;
char *expected_output;
};
int main() {
int nest = 0;
int TESTSEP = 0;
char *temp_malloc_string;
if (TESTSEP>0) {
char *tokens = "# this is a comment\n{nesting {example}
unnesting}\n \t end";
temp_malloc_string = malloc(strlen(tokens)*sizeof(char));
strcpy(temp_malloc_string, tokens);
char * t = token(&nest, &temp_malloc_string);
printf("nest: %d\n", nest);
printf("tokens: %s\n", t);
separator(&nest, &temp_malloc_string);
printf("nest: %d\n", nest);
printf("tokens: %s\n", temp_malloc_string);
return 0;
} else {
struct test_case test_cases[] = {
{"hello world", 0, "hello"},
{"hello#world", 0, "hello"},
{"hello{world}", 0, "hello"},
{"hello world", 0, "hello"},
{"hello\tworld", 0, "hello"},
{"hello\nworld", 0, "hello"},
};
for (int i = 0; i < sizeof(test_cases) / sizeof(test_cases[0]); i++) {
struct test_case test_case = test_cases[i];
char *tokens = test_case.input;
printf("len of string is %d\n", strlen(tokens));
temp_malloc_string = malloc((strlen(tokens)+1)*sizeof(char));
char * tt = temp_malloc_string;
if ( temp_malloc_string==NULL ) {
printf("error!\n");
}
strcpy(temp_malloc_string, tokens);
printf("tm going in: %s\n", temp_malloc_string);
char *output = token(&test_case.nest, &temp_malloc_string);
printf("Test case %d: expected %s, got %s\n\t\ttm is now: %s\n",
i, test_case.expected_output, output, temp_malloc_string);
if (strcmp(output, test_case.expected_output) != 0) {
printf("Test case %d failed: expected %s, got %s\n",
i, test_case.expected_output, output);
}
free(tt);
temp_malloc_string = NULL;
}
return 0;
}
}
Now when I run the code I get:
./explore.exe
len of string is 11
tm going in: hello world
Test case 0: expected hello, got hello
tm is now: world
len of string is 11
tm going in: hello#world
Test case 1: expected hello, got hello
tm is now:
len of string is 12
tm going in: hello{world}
Test case 2: expected hello, got hello
tm is now: world}
len of string is 12
tm going in: hello world
Test case 3: expected hello, got hello
tm is now: world
len of string is 11
tm going in: hello world
Test case 4: expected hello, got hello
tm is now: world
len of string is 11
tm going in: hello
world
Test case 5: expected hello, got hello
tm is now: world
Process finished with exit code 0
And when I stop at the breakpoint I can write to memory.
In this modified code I malloc a char* object and copy the string from the struct into that then pass that into the token function.
I'm guess that gdb is protecting me from writing to the .text block in code.
Like I said: I'm a newbie :(
I have a long string of about 50,000,000 long... , and I am substituting it part by part
cat FILE | tail -n+2 | awk -v k=100 '{
i = 1
while (i<length($0)-k+1) {
x = substr($0, i, k)
if (CONDITION) {
x changed sth
$0 = substr($0,1,i-1) x substr($0,i+k)
}
i += 1
}
gsub(sth,sth,$0)
printf("%s",$0) >> FILE
}'
Are there any ways to replace $0 at position i with x of length k without using this method?
The string is too long and the commands runs extremely slow
sample input:
NNNNNNNNNNggcaaacagaatccagcagcacatcaaaaagcttatccacAGTAATTCATTATATCAAAATGCTCCAggccaggcgtggtggcttatgcc
sample output:
NNNNNNNNNNggcnnncngnnnccngcngcncnncnnnnngcnnnnccncNGNNNNNCNNNNNNNCNNNNNGCNCCNggccnggcgnggnggcnnnngcc
If substring with length k=10 contains >50% of A || a || T || t
(so there are length($0)-k+1 substrings)
substitute A and T with N, a and t with n
The $0 string must maintain it size and sequence (Case sensitive)
EDIT:
I misunderstood the requirement of this problem, and repost the question at here.
Basically:
read a window of characters to two buffers - scratch buffer and output buffer
if in the scratch buffer there are more then some count of characters ATat
then replace all characters ATat in the output buffer buffer to Nn respectively
output one character from the output buffer
flush one character in both buffers
and go to step 1 to repeat reading the characters into buffers
when the end of line is encountered, just flush output buffer and reset it all
A small C program for sure is going to be the fastest:
// The window size
#define N 10
// The percent of the window that has to be equal to one of [AaTt]
#define PERCENT 50
#include <assert.h>
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
// output a string
static void output(char *outme, size_t n) {
fwrite(outme, n, 1, stdout);
}
// is one of [AaTt]
static bool is_one_of_them(char c) {
switch(c) {
case 'A':
case 'a':
case 'T':
case 't':
return true;
}
return false;
}
// Convert one of characters to n/N depending on case
static char convert_them_to_n(char c) {
// switch(c){ case 'T': case 'A': return true; } return false;
// ASCII is assumed
const char m = ~0x1f;
const char w = 'n' & ~m;
return (c & m) | w;
}
static const unsigned threshold = N * PERCENT / 100;
// Store the input in buf
static char buf[N];
// Store the output to-be-outputted in out
static char out[N];
// The current position in buf and out
// The count of readed characters
static size_t pos;
// The count of one of searched characters in buf
static unsigned count_them;
static void buf_reset(void) {
pos = 0;
count_them = 0;
}
static void buf_flush(void) {
output(out, pos);
buf_reset();
}
static void buf_replace_them(void) {
// TODO: this could keep count of characters alrady replaced in out to save CPU
for (size_t i = 0; i < N; ++i) {
if (is_one_of_them(out[i])) {
out[i] = convert_them_to_n(out[i]);
}
}
}
static void buf_flush_one(void) {
assert(pos > 0);
assert(pos == N);
output(out, 1);
count_them -= is_one_of_them(buf[0]);
memmove(buf, buf + 1, pos - 1);
memmove(out, out + 1, pos - 1);
pos--;
}
static void buf_add(char c) {
buf[pos] = out[pos] = c;
pos++;
count_them += is_one_of_them(c);
// if we reached the substring length
if (pos == N) {
// if the count reached the threshold
if (count_them >= threshold) {
// convert the characters to n
buf_replace_them();
}
// flush one character only at a time
buf_flush_one();
}
}
int main() {
int c;
buf_reset();
while ((c = getchar()) != EOF) {
if (c == '\n') {
// If its a newline, just flush what we have buffered
buf_flush();
output("\n", 1);
continue;
}
buf_add(c);
}
buf_flush();
}
Such a C program is easily transferable to for example an awk script, just one need to read one character at a time. Below I split the characters with split, like:
awk -v N=10 -v percent=50 '
BEGIN{ threshold = N * percent / 100; pos=0 }
function is_one_of_them(c) {
return c ~ /^[aAtT]$/;
}
function buf_flush(i) {
for (i = 0; i < pos; ++i) {
printf "%s", out[i]
}
pos = 0
count_them = 0
}
function buf_replace_them(i) {
for (i = 0; i < pos; ++i) {
if (is_one_of_them(out[i])) {
out[i] = out[i] ~ /[AT]/ ? "N" : "n";
}
}
}
function buf_flush_one(i) {
printf "%s", out[0]
count_them -= is_one_of_them(buf[0])
if(0 && debug) {
printf(" count_them %s ", count_them)
for (i = 0; i < pos-1; ++i) {
printf("%s", buf[i+1])
} printf(" ");
for (i = 0; i < pos-1; ++i) {
printf("%s", out[i+1])
}
printf("\n");
}
for (i = 0; i < pos-1; ++i) {
buf[i] = buf[i+1]
out[i] = out[i+1]
}
pos--
}
function buf_add(c) {
buf[pos]=c; out[pos]=c; pos++
count_them += is_one_of_them(c)
if (pos == N) {
if (count_them >= threshold) {
buf_replace_them()
}
buf_flush_one()
}
}
{
split($0, chars, "")
for (idx = 0; idx <= length($0); idx++) {
buf_add(chars[idx])
}
buf_flush();
printf "\n";
}
'
Both programs when run with the input presented in the first line produce the output presented in the second line (note that lone a near the end is not replaced, because there are no 5 charactets ATat in a window of 10 characters from it):
NNNNNNNNNNggcaaacagaatccagcagcacatcaaaaagcttatccacAGTAATTCATTATATCAAAATGCTCCAggccaggcgtggtggcttatgcc
NNNNNNNNNNggcnnncngnnnccngcngcncnncnnnnngcnnnnccncNGNNNNNCNNNNNNNCNNNNNGCNCCNggccaggcgnggnggcnnnngcc
Both solutions were tested on repl.
You need to be careful with how you address this problem. You cannot work on the substituted string. You need to keep track of the original string. Here is a simple example. Assume we have a string consisting of x and y and we want to replace all y with z if there are 8 y in a substring of 10. Imagine your input looks like:
yyyyyyyyxxy
The first substring of 10 reads yyyyyyyyxx and would be translated into zzzzzzzzxx. If you perform the substitution directly into the original string, you get zzzzzzzzxxy. The second substring now reads zzzzzzzxxy, and does not contain 8 times y, while in the original string it does. So according to the solution of the OP, this would lead into inconsistent results, depending on if you start from the front or the back. So a quick solution would be:
awk -v N=10 -v p=50 '
BEGIN { n = N*p/100 }
{ s = $0 }
{ for(i=1;i<=length-N;++i) {
str=substr($0,i,N)
c=gsub(/[AT]/,"N",str) + gsub(/[at]/,"n",str)
if(c >= n) s = substr(s,1,i-1) str substr(s,i+N)
}
}
{ print s }' file
There is ofcourse quite some work you do double here. Imagine you have a string of the form xxyyyyyyyyxx, you would perform 4 concatinations while you only need to do one. So the best idea is to minimalise the work and only check the substrings which end with the respective character:
awk -v N=10 -v p=50 '
BEGIN { n = N*p/100 }
{ s = $0 }
{ i=N; while (match(substr($0,i),/[ATat]/)) {
str=substr($0,i+RSTART-N,N)
c=gsub(/[AT]/,"N",str) + gsub(/[at]/,"n",str)
if(c >= n) { s = substr(s,1,i+RSTART-N-1) str substr(s,i+RSTART)}
i=i+RSTART
}
}
{ print s }' file
To replace $0 at position i with x do:
awk 'BEGIN{i=12345;x="blubber"}
{
printf("%s",substr($0,1,i));
printf("%s",x);
printf("%s",substr($0,i+length(x)));
}'
I don't think there is any faster method.
To replace AGCT with N and agct with n use tr. To replace them only within a range and using awk you should do:
awk 'BEGIN{i=12345;n=123}
{
printf("%s",substr($0,1,i-1));
printf(gsub(/[atgc]/,"n",gsub(/[ATGC]/,"N",substr($0,i,i+n-1))));
printf("%s",substr($0,i+n));
}'
To do more advanced and faster processing you should consider c/c++.
I created a function that compares strings, and as I was frustrated about it always missing the last character in the second string and always returning "identical strings" as a result, I noticed that I was messing around and used gets() instead of fgets() for the second string. I changed that and the function works as expected.
My question is, why does the gets() function subtract that last character? Shouldn't it subtract the null and leave it at that?
Does that mean that as a newcomer to C, I should avoid using gets() and focus on fgets() instead? I'm starting to think of them in the same way I think of strcmp() vs strncmp()
Thanks for your time everyone!
Note: I'm aware that I don't really need the (i==j) at the end, I just left it there (extra security, maybe?).
bool compare_string(const char *string1, const char *string2) {
int i = 0, j = 0, result = 0;
while (string1[i] != '\0') {
i++;
}
while (string2[j] != '\0') {
j++;
}
i = 0;
j = 0;
while ((string1[i] != '\0') && (string2[j] != '\0')) {
if (string1[i] < string2[j]) {
result = -1;
break;
} else if (string1[i] > string2[j]) {
result = 1;
break;
} else if (string1[i] == string2[j]) {
result = 0;
}
i++;
j++;
}
if ((result == 0) && (i==j)) {
printf("identical strings \n");
} else if (result == -1) {
printf("not identical, -1 \n");
} else if (result == 1) {
printf("not identical, 1 \n");
}
}
//in main
char str_compare1[STRING_LIMIT];
char str_compare2[STRING_LIMIT];
printf("enter 1st string to compare, (100) characters or less: \n");
fgets(str_compare1, STRING_LIMIT, stdin);
printf("enter 2nd string to compare, (100) characters or less \n");
fgets(str_compare2, STRING_LIMIT, stdin);
result = compare_string(str_compare1, str_compare2);
I just saw that this could technically work, the only mistake I couldn´t resolve was the last ASCII character that gets printed everytime I test it out, I also tested this out without using the name variable, I mean just making a substraction of 32 to any lower case letter in ASCII should give me their upper case one and it does, but I´m curious on why I´m getting an additional char, wich from what I see in screen is apparently Û.
#include <stdio.h>
main()
{
char name[22];
int i;
fputs("Type your name ",stdout);
fgets(name,22,stdin);
for (i = 0; name[i] != '\0'; i = i + 1)
printf("%c",(name[i])-32); /*This will convert lower case to upper */
/* using as reference the ASCII table*/
fflush(stdin);
getchar();
}
Perhaps there is a line break character at the end of the string.
You can check the chararacter code, so that you only convert characters that actually are lower case letters:
for (i = 0; name[i] != '\0'; i = i + 1) {
char c = name[i];
if (c => 97 && c <= 122) {
c -= 32;
}
printf("%c", c);
}
void read_chararray(char in_array[], int* Length)
{
int Indx = 0, Indx2 = 0, Indx3 = 0; // int declarations for indexs of some loops
char cinput = { 0 }, word[255] = { 0 }, word2[255] = { 0 }; // declaration of cinput and first char array before punctiation removed
for (Indx = 0; (cinput = getchar()) != '\n'; Indx++) { // Loop for getting characters from user stop at <enter>
word[Indx] = cinput; // Placing char into array while changing to lowercase
}
Indx2 = Indx; // Set Indx2 to Indx for loop operation
for (Indx = 0; Indx < Indx2; Indx++) { // Loop to check and replace upper characters with lower
cinput = word[Indx];
if (cinput >= 65 && cinput <= 90) { // If cinput is within the ASCII range 65 and 90, this indicates upper characters
cinput += 32; // Add 32 to cinput to shift to the lower character range within the ASCII table
in_array[Indx] = cinput; // Input new value into array pointer
}
else if (cinput >= 97 && cinput <= 122) // scans if character are lower ASCII, places them in array irraticating punctuation and whitespce
in_array[Indx] = cinput; // Input remaining lower case into array pointer
}
*Length = Indx; // final size of array set to Length variable for future use
}
#include<stdio.h>
void upper(char);
void main()
{
char ch;
printf("\nEnter the character in lower case");
scanf("%c", &ch);
upper(ch);
}
void upper( char c)
{
printf("\nUpper Case: %c", c-32);
}
I wanted to make a custom ITOA function to put large numbers into small strings, this is what I have coded :
main(){
printf("itoa(2000000000,36)= '%s'",itoa(2000000000,36));
printf("itoa(36,36)= '%s'",itoa(36,36));
printf("itoa(37,36)= '%s'",itoa(37,36));
return 1;
}
stock itoa(val, base)
{
new buf[1024] = {0,...};
new i = 1023;
new LETTERZ[37] = {'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z',0};
for(; val && i; --i, val /= base)
buf[i] = LETTERZ[val % base];
return buf[i+1];
}
It's based on 'C' code from this page: http://www.jb.man.ac.uk/~slowe/cpp/itoa.html
But somehow this is the output:
[20:34:35] itoa(2000000000,36)= 'X'
[20:34:35] itoa(36,36)= '1'
[20:34:35] itoa(37,36)= '1'
And this is totally wrong, I don't know which output to expect but 36 and 37 for sure can't be the same output and 2 000 000 000 can't be just 'X', as X is suposed to be 35, not 2 000 000 000,
ZZ should be 1295 I think... I want to base this on the hexadecimal system, but with all the alfabet letters.
Could anyone tell me what's wrong here?
I'm working with a typeless language called PAWN (also known as SMALL) and later i want to use this code in VB.NET
/* itoa example */
#include <stdio.h>
#include <stdlib.h>
int main ()
{
int i;
char buffer [33];
printf ("Enter a number: ");
scanf ("%d",&i);
itoa (i,buffer,10);
printf ("decimal: %s\n",buffer);
itoa (i,buffer,16);
printf ("hexadecimal: %s\n",buffer);
itoa (i,buffer,2);
printf ("binary: %s\n",buffer);
return 0;
}
You only give the number and the base, but parameter 2 needs a pointer to char already allocated. Use a buffer or try NULL, so the function will return the result.
THe solution seemed to be simple, the return buf[i+1] just returned one character so what I did is make it return an array:
new _s#T[4096];
#define sprintf(%1) (format(_s#T, SPRINTF_MAX_STRING, %1), _s#T)
main(){
new num = atoi("ABCDEFG",36);
printf("%d",num);
printf("%s",itoa(num,36));
return 1;
}
stock itoa(val, base)
{
new buf[1024] = {0,...};
new LETTERZ[37] = {'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z',0};
for(new pos = 0; val;++pos,val = floatround(val/base,floatround_floor))
strins(buf,sprintf("%c",LETTERZ[val % base]),0);
return buf;
}
stock atoi(val[], base)
{
new CURRNUM = 0;
new len = strlen(val);
new LETTERZ[37] = {'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z',0};
for(new i = 0; i < len; ++i)
{
for(new x = 0; x < base; ++x)
{
new y = (len-i)-1;
if(val[y] == LETTERZ[x])
{
CURRNUM += x*floatround(floatpower(base,i));
}
}
}
return CURRNUM;
}