I encountered this ascii's style ascii table.
Of course I can store it in a file ascii and use cat ascii to display it content.
But I want to make it behavior more like a command.
UPDATE
When I read cs:app I find that how I bother to restore it in a file and using other commands.
Just run man ascii
If your shell supports aliases, you can do:
alias ascii='cat ~/ascii'
Then just type ascii et voila!
If you're using bash, put the above line in your .bashrc to persist it across logins. Other shells have similar features.
Dec Hex Dec Hex Dec Hex Dec Hex Dec Hex Dec Hex Dec Hex Dec Hex
0 00 NUL 16 10 DLE 32 20 48 30 0 64 40 # 80 50 P 96 60 ` 112 70 p
1 01 SOH 17 11 DC1 33 21 ! 49 31 1 65 41 A 81 51 Q 97 61 a 113 71 q
2 02 STX 18 12 DC2 34 22 " 50 32 2 66 42 B 82 52 R 98 62 b 114 72 r
3 03 ETX 19 13 DC3 35 23 # 51 33 3 67 43 C 83 53 S 99 63 c 115 73 s
4 04 EOT 20 14 DC4 36 24 $ 52 34 4 68 44 D 84 54 T 100 64 d 116 74 t
5 05 ENQ 21 15 NAK 37 25 % 53 35 5 69 45 E 85 55 U 101 65 e 117 75 u
6 06 ACK 22 16 SYN 38 26 & 54 36 6 70 46 F 86 56 V 102 66 f 118 76 v
7 07 BEL 23 17 ETB 39 27 ' 55 37 7 71 47 G 87 57 W 103 67 g 119 77 w
8 08 BS 24 18 CAN 40 28 ( 56 38 8 72 48 H 88 58 X 104 68 h 120 78 x
9 09 HT 25 19 EM 41 29 ) 57 39 9 73 49 I 89 59 Y 105 69 i 121 79 y
10 0A LF 26 1A SUB 42 2A * 58 3A : 74 4A J 90 5A Z 106 6A j 122 7A z
11 0B VT 27 1B ESC 43 2B + 59 3B ; 75 4B K 91 5B [ 107 6B k 123 7B {
12 0C FF 28 1C FS 44 2C , 60 3C < 76 4C L 92 5C \ 108 6C l 124 7C |
13 0D CR 29 1D GS 45 2D - 61 3D = 77 4D M 93 5D ] 109 6D m 125 7D }
14 0E SO 30 1E RS 46 2E . 62 3E > 78 4E N 94 5E ^ 110 6E n 126 7E ~
15 0F SI 31 1F US 47 2F / 63 3F ? 79 4F O 95 5F _ 111 6F o 127 7F DEL
Related
I have a csv file /tmp/test.csv with the following content.
VM,Datacenter,Cluster,Host,Folder,OS,VM ID,VM UUID,vCenter UUID
A0F0US014XVM022,"/AMMWDC04_DC/A0F0 <96> AA DR","Red Hat Enterprise Linux 6 (64-bit)",vm-2910,421f2eba-8b60-6166-3b56-f22e3f71eecf,94694731-df3a-4ee6-9962-49df97a6f08d
I want to replace <96> (surrounded by space) with - in the csv file. I tried sed -i -e 's/<96>/-/g' /tmp/test.csv but this did not work. May be because of the special symbols involved.
sed version 4.2.1
[root#fmsprdchef001 ~]# grep vm-2910 /tmp/test.csv | hexdump -C
00000000 41 30 46 30 55 53 30 31 34 58 56 4d 30 32 32 2c |A0F0US014XVM022,|
000000e0 41 4d 4d 57 44 43 30 34 5f 44 43 2f 41 30 46 30 |AMMWDC04_DC/A0F0|
000000f0 20 96 20 41 41 20 44 52 22 2c 22 52 65 64 20 48 | . AA DR","Red H|
00000100 61 74 20 45 6e 74 65 72 70 72 69 73 65 20 4c 69 |at Enterprise Li|
00000110 6e 75 78 20 36 20 28 36 34 2d 62 69 74 29 22 2c |nux 6 (64-bit)",|
00000120 76 6d 2d 32 39 31 30 2c 34 32 31 66 32 65 62 61 |vm-2910,421f2eba|
00000130 2d 38 62 36 30 2d 36 31 36 36 2d 33 62 35 36 2d |-8b60-6166-3b56-|
00000140 66 32 32 65 33 66 37 31 65 65 63 66 2c 39 34 36 |f22e3f71eecf,946|
00000150 39 34 37 33 31 2d 64 66 33 61 2d 34 65 65 36 2d |94731-df3a-4ee6-|
00000160 39 34 36 32 2d 34 39 64 66 39 37 61 36 66 30 38 |9462-49df97a6f08|
00000170 64 0a 41 30 46 30 55 53 30 31 34 58 56 4d 30 32 |d.A0F0US014XVM02|
00000180 32 2c 70 6f 77 65 72 65 64 4f 6e 2c 46 61 6c 73 |2,poweredOn,Fals|
00000190 65 2c 56 6d 78 6e 65 74 33 2c 2c 2c 54 72 75 65 |e,Vmxnet3,,,True|
000001a0 2c 54 72 75 65 2c 30 30 3a 35 30 3a 35 36 3a 39 |,True,00:50:56:9|
000001b0 66 3a 30 31 3a 62 38 2c 61 73 73 69 67 6e 65 64 |f:01:b8,assigned|
000001c0 2c 22 31 30 2e 31 30 30 2e 31 2e 31 32 2c 20 66 |,"10.100.1.12, f|
000001d0 65 38 30 3a 3a 32 35 30 3a 35 36 66 66 3a 66 65 |e80::250:56ff:fe|
000001e0 39 66 3a 31 62 38 22 2c 22 41 6d 65 72 69 63 61 |9f:1b8","America|
000001f0 6e 20 41 69 72 6c 69 6e 65 73 3b 20 50 52 44 3b |n Airlines; PRD;|
00000200 20 61 70 70 20 26 20 44 42 32 3b 44 52 22 2c 41 | app & DB2;DR",A|
00000210 4d 4d 57 44 43 30 34 5f 44 43 2c 41 4d 4d 57 44 |MMWDC04_DC,AMMWD|
00000220 43 30 34 43 41 2c 61 6d 6d 77 64 63 30 34 63 75 |C04CA,ammwdc04cu|
00000230 73 74 65 73 78 30 31 2e 69 6d 7a 63 6c 6f 75 64 |stesx01.imzcloud|
00000240 2e 69 62 6d 61 6d 6d 73 61 70 2e 6c 6f 63 61 6c |.ibmammsap.local|
00000250 2c 22 2f 41 4d 4d 57 44 43 30 34 5f 44 43 2f 41 |,"/AMMWDC04_DC/A|
00000260 30 46 30 20 96 20 41 41 20 44 52 22 2c 22 52 65 |0F0 . AA DR","Re|
00000270 64 20 48 61 74 20 45 6e 74 65 72 70 72 69 73 65 |d Hat Enterprise|
00000280 20 4c 69 6e 75 78 20 36 20 28 36 34 2d 62 69 74 | Linux 6 (64-bit|
00000290 29 22 2c 76 6d 2d 32 39 31 30 2c 34 32 31 66 32 |)",vm-2910,421f2|
000002a0 65 62 61 2d 38 62 36 30 2d 36 31 36 36 2d 33 62 |eba-8b60-6166-3b|
000002b0 35 36 2d 66 32 32 65 33 66 37 31 65 65 63 66 2c |56-f22e3f71eecf,|
000002c0 39 34 36 39 34 37 33 31 2d 64 66 33 61 2d 34 65 |94694731-df3a-4e|
000002d0 65 36 2d 39 34 36 32 2d 34 39 64 66 39 37 61 36 |e6-9462-49df97a6|
000002e0 66 30 38 64 0a |f08d.|
000002e5
This is the important part from your hexdump:
000000f0 20 96 20 41 41 20 44 52 22 2c 22 52 65 64 20 48 | . AA DR","Red H|
I suggest:
sed -i 's/ \x96 /-/' file
Following sed may help you on same.
sed 's/\([^<]*\)\(<.*>\)\(.*\)/\1-\3/' Input_file
In case you want to save the output into same Input_file then use -i option with sed in above code.
EDIT: In case your Input_file is having control characters then use following.
sed 's/\r//g;s/\([^<]*\)\(<.*>\)\(.*\)/\1-\3/' Input_file
I am using Net::Telnet to connect to a HP ProCurve Switch to login and backup the config. However I ran into issues where waitfor returns VT100/ANSI control chars:
< 0x00000: ff fd 18 ff fd 1f ff fb 01 1b 5b 32 4a 1b 5b 3f ..........[2J.[?
< 0x00010: 37 6c 1b 5b 33 3b 32 33 72 1b 5b 3f 36 6c 1b 5b 7l.[3;23r.[?6l.[
< 0x00020: 31 3b 31 48 1b 5b 3f 32 35 6c 1b 5b 31 3b 31 48 1;1H.[?25l.[1;1H
< 0x00030: 48 50 20 4a 39 37 32 38 41 20 32 39 32 30 2d 34 HP J9728A 2920-4
< 0x00040: 38 47 20 53 77 69 74 63 68 0d 0d 0a 53 6f 66 74 8G Switch...Soft
< 0x00050: 77 61 72 65 20 72 65 76 69 73 69 6f 6e 20 57 42 ware revision WB
< 0x00060: 2e 31 35 2e 31 32 2e 30 30 31 35 0d 0d 0a 0d 0d .15.12.0015.....
< 0x00070: 0a 43 6f 70 79 72 69 67 68 74 20 28 43 29 20 31 .Copyright (C) 1
< 0x00080: 39 39 31 2d 32 30 31 34 20 48 65 77 6c 65 74 74 991-2014 Hewlett
< 0x00090: 2d 50 61 63 6b 61 72 64 20 44 65 76 65 6c 6f 70 -Packard Develop
< 0x000a0: 6d 65 6e 74 20 43 6f 6d 70 61 6e 79 2c 20 4c 2e ment Company, L.
< 0x000b0: 50 2e 0d 0a 0d 0a 20 20 20 20 20 20 20 20 20 20 P.....
Unfortunately, this screws up waitfor because if I try to waitfor(/^password:/i) it will return a string with those control chars in it, or wait forever since the regex is never matched.
Is there any way to have Net::Telnet automatically remove those control characters? Is there any way to have waitfor only care about ASCII printable characters?
Consider the following script.
#!/bin/bash
echo {00..99}
n=99
echo {00..$n}
The output of this script is:
00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99
{00..99}
The desired output is:
00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99
00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99
One solution which produces the desired output is
eval echo {00..$n}
Unfortunately, this solution uses eval which I'd prefer to avoid if possible.
Does anyone know of a way to obtain the desired result using brace expansion but not eval?
From the bash manual:
The order of expansions is: brace expansion; tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion); word splitting; and pathname expansion.
Given that variable expansion comes after brace expansion, and that there is no way to induce a different order of operations without using eval, I would have to conclude that no, there is no way to avoid using eval.
Does anyone know of a way to obtain the desired without without eval?
You can use seq command,
seq -w -s ' ' 0 $n
Test:
sat $ seq -w -s " " 0 $n
00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99
Not sure if this meets your requirements, as it doesn't use braces, but (with GNU seq at least) the following produces the desired output:
$ n=99
$ seq -f%02.0f -s' ' 00 "$n"
00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99
The "-f" option produces the zero-padding, and the "-d" uses spaces to separate, rather than newlines.
I am trying to solve the Projecteuler #11 but I am running into an error when I'm trying to create a function to calculate the multiplication of every 4 numbers in a column. I am getting an error:
Project11.rb:59:in `sumvertical': undefined method `[]' for nil:NilClass (NoMeth
odError)
I feel like there is something I am easily overlooking here. I appreciate the help!
#project #11 http://projecteuler.net/problem=11
grid="08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48"
grid=grid.split()
grid=grid.collect {|s| s.to_i}
multiarray=[]
i = 0
e = 19
until e > 400
multiarray << grid[i..e]
i+= 20
e+= 20
end
def sumhorizontal(x) #checks sum of all horizontal 4 elements
sum = 0
x.each {|a|
i=0
e=3
while e < a.length
if a[i..e].inject(:*) > sum
sum = a[i..e].inject(:*)
i += 1
e += 1
else
i += 1
e += 1
end
end
}
return sum
end
def sumvertical(x)
sum = 0
i=0
e=0
while e < x.length #Will break once the end point is longer than the length of an array
until i > 20 #Checks the first column
if x[i][e]*x[i+1][e]*x[i+2][e]*x[i+3][e] > sum #Error is here
sum = x[i][e]*x[i+1][e]*x[i+2][e]*x[i+3][e]
i += 1
else
i += 1
end
end
e += 1 #once you are out of the until statement, it increases e by 1 to check the next column
i = 0 #resets i so it can go back to the zero
end
return sum
end
print sumvertical(multiarray)
The grid has 20 rows. Your loop is actually trying to reach all the way to a 24rd row; that's because it goes through 21 iterations (i starts at 0, and goes until it equals 21), and each iteration reaches 3 beyond the current value of i (when you call x[i+3]). When i is 17, your code will break, because x[i+3][e] is trying to index into the 21st row of x. i+3 is 20, but the highest available index is 19. So what happens is, x[20] returns nil, and then the [] method is called on nil, which generates your error.
Also, the standard library has a transpose method that you can call on your array. If you use it, you just need one method (sumhorizontal). You can get the column sums with sumhorizontal(multiarray.transpose).
One more thing... it looks like you're coming from a procedural language. Ruby has an extensive standard library and coding constructs that can save you a lot of time and keystrokes. There is typically no need to iterate with while loops and index variables in Ruby. sumhorizontal, for instance, can be written like this (it should really be called producthorizontal, though if you're trying to solve Project Euler #11:
def sumhorizontal(x)
x.map { |r| r.each_slice(4).map { |s| s.reduce(:*) }.max }.max
end
Good luck with the rest of your Ruby learning journey!
I found that wcslen() in VC++2010 returns correct count of letters; meanwhile Xcode does not.
For example, the code below returns correct 11 in VC++ 2010, but returns incorrect 17 in Xcode 4.2.
const wchar_t *p = L"123abc가1나1다";
size_t plen = wcslen(p);
I guess Xcode app stores wchar_t string as UTF-8 in memory. This is another strange thing.
How can I get 11 just like VC++ in Xcode too?
I ran this program on a Mac Mini running MacOS X 10.7.2 (Xcode 4.2):
#include <stdio.h>
#include <wchar.h>
int main(void)
{
const wchar_t p[] = L"123abc가1나1다";
size_t plen = wcslen(p);
if (fwide(stdout, 1) <= 0)
{
fprintf(stderr, "Failed to make stdout wide-oriented\n");
return -1;
}
wprintf(L"String <<%ls>>\n", p);
putwc(L'\n', stdout);
wprintf(L"Length = %zu\n", plen);
for (size_t i = 0; i < sizeof(p)/sizeof(*p); i++)
wprintf(L"Character %zu = 0x%X\n", i, p[i]);
return 0;
}
When I do a hex dump of the source file, I see:
0x0000: 23 69 6E 63 6C 75 64 65 20 3C 73 74 64 69 6F 2E #include <stdio.
0x0010: 68 3E 0A 23 69 6E 63 6C 75 64 65 20 3C 77 63 68 h>.#include <wch
0x0020: 61 72 2E 68 3E 0A 0A 69 6E 74 20 6D 61 69 6E 28 ar.h>..int main(
0x0030: 76 6F 69 64 29 0A 7B 0A 20 20 20 20 63 6F 6E 73 void).{. cons
0x0040: 74 20 77 63 68 61 72 5F 74 20 70 5B 5D 20 3D 20 t wchar_t p[] =
0x0050: 4C 22 31 32 33 61 62 63 EA B0 80 31 EB 82 98 31 L"123abc...1...1
0x0060: EB 8B A4 22 3B 0A 20 20 20 20 73 69 7A 65 5F 74 ...";. size_t
0x0070: 20 70 6C 65 6E 20 3D 20 77 63 73 6C 65 6E 28 70 plen = wcslen(p
0x0080: 29 3B 0A 20 20 20 20 69 66 20 28 66 77 69 64 65 );. if (fwide
0x0090: 28 73 74 64 6F 75 74 2C 20 31 29 20 3C 3D 20 30 (stdout, 1) <= 0
0x00A0: 29 0A 20 20 20 20 7B 0A 20 20 20 20 20 20 20 20 ). {.
0x00B0: 66 70 72 69 6E 74 66 28 73 74 64 65 72 72 2C 20 fprintf(stderr,
0x00C0: 22 46 61 69 6C 65 64 20 74 6F 20 6D 61 6B 65 20 "Failed to make
0x00D0: 73 74 64 6F 75 74 20 77 69 64 65 2D 6F 72 69 65 stdout wide-orie
0x00E0: 6E 74 65 64 5C 6E 22 29 3B 0A 20 20 20 20 20 20 nted\n");.
0x00F0: 20 20 72 65 74 75 72 6E 20 2D 31 3B 0A 20 20 20 return -1;.
0x0100: 20 7D 0A 20 20 20 20 77 70 72 69 6E 74 66 28 4C }. wprintf(L
0x0110: 22 53 74 72 69 6E 67 20 3C 3C 25 6C 73 3E 3E 5C "String <<%ls>>\
0x0120: 6E 22 2C 20 70 29 3B 0A 20 20 20 20 70 75 74 77 n", p);. putw
0x0130: 63 28 4C 27 5C 6E 27 2C 20 73 74 64 6F 75 74 29 c(L'\n', stdout)
0x0140: 3B 0A 20 20 20 20 77 70 72 69 6E 74 66 28 4C 22 ;. wprintf(L"
0x0150: 4C 65 6E 67 74 68 20 3D 20 25 7A 75 5C 6E 22 2C Length = %zu\n",
0x0160: 20 70 6C 65 6E 29 3B 0A 20 20 20 20 66 6F 72 20 plen);. for
0x0170: 28 73 69 7A 65 5F 74 20 69 20 3D 20 30 3B 20 69 (size_t i = 0; i
0x0180: 20 3C 20 73 69 7A 65 6F 66 28 70 29 2F 73 69 7A < sizeof(p)/siz
0x0190: 65 6F 66 28 2A 70 29 3B 20 69 2B 2B 29 0A 20 20 eof(*p); i++).
0x01A0: 20 20 20 20 20 20 77 70 72 69 6E 74 66 28 4C 22 wprintf(L"
0x01B0: 43 68 61 72 61 63 74 65 72 20 25 7A 75 20 3D 20 Character %zu =
0x01C0: 30 78 25 58 5C 6E 22 2C 20 69 2C 20 70 5B 69 5D 0x%X\n", i, p[i]
0x01D0: 29 3B 0A 20 20 20 20 72 65 74 75 72 6E 20 30 3B );. return 0;
0x01E0: 0A 7D 0A .}.
0x01E3:
The output when compiled with GCC is:
String <<123abc
Length = 11
Character 0 = 0x31
Character 1 = 0x32
Character 2 = 0x33
Character 3 = 0x61
Character 4 = 0x62
Character 5 = 0x63
Character 6 = 0xAC00
Character 7 = 0x31
Character 8 = 0xB098
Character 9 = 0x31
Character 10 = 0xB2E4
Character 11 = 0x0
Note that the string is truncated at the zero byte - I think that is probably a bug in the system, but it seems a little unlikely that I'd manage to find one on my first attempt at using wprintf(), so it is more likely I'm doing something wrong.
You're right, in the multi-byte UTF-8 source code, the string occupies 17 bytes (8 one-byte basic Latin-1 characters, and 3 characters each encoded using 3 bytes). So, the raw strlen() on the source string would return 17 bytes.
GCC version is:
i686-apple-darwin11-llvm-gcc-4.2 (GCC) 4.2.1 (Based on Apple Inc. build 5658) (LLVM build 2335.15.00)
Copyright (C) 2007 Free Software Foundation, Inc.
This is free software; see the source for copying conditions. There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
Just for giggles, I tried clang, and I get a different result. Compiled using:
clang -o row row.c -Wall -std=c99
using:
Apple clang version 2.1 (tags/Apple/clang-163.7.1) (based on LLVM 3.0svn)
Target: x86_64-apple-darwin11.3.0
Thread model: posix
The output when compiled with clang is:
String <<123abc가1나1다>>
Length = 17
Character 0 = 0x31
Character 1 = 0x32
Character 2 = 0x33
Character 3 = 0x61
Character 4 = 0x62
Character 5 = 0x63
Character 6 = 0xEA
Character 7 = 0xB0
Character 8 = 0x80
Character 9 = 0x31
Character 10 = 0xEB
Character 11 = 0x82
Character 12 = 0x98
Character 13 = 0x31
Character 14 = 0xEB
Character 15 = 0x8B
Character 16 = 0xA4
Character 17 = 0x0
So, now the string appears correctly, but the length is given as 17 instead of 11. Superficially, you can take your choice of bugs - string looks OK (in a terminal - /Applications/Utilities/Terminal - acclimatized to UTF8) but length is wrong, or length is right but string does not appear correctly.
I note that sizeof(wchar_t) in both gcc and clang is 4.
The left hand does not understand what the right hand is doing. I think there's a case for claiming both are broken, in different ways.