Thread: strlen with escape codes

having got my head around arrays (for the next 4 seconds till i go thick again) i am wondering about the return value from strlen

Code:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main()
{
    char mystring[40] = "hello";

    printf("the length of \"my string\" is %lu\n", strlen(mystring));

    return 0;
}

this returns 5 as one would expect (5 letters but doesn't count the "\0" character).

however if i do this:

Code:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main()
{
    char mystring[40] = "\e[196;5;196m\e[38;5;196mhello\e[0m";

    printf("the length of \"my string\" is %lu\n", strlen(mystring));

    return 0;
}

i count 35 characters but lenstr returns 32. if its ignoring the \e parts i make the count 29 so it cant be that. what has happened to the "missing" 3 characters and when deciding on the length of an array to hold these escape codes and message what do i declare the array length as 36 or 33 including the null.

many thanks
coop

"\e[192;5;196m" (12 chars)
"\e[38;5;196m" (11 chars)
"hello" (5 chars)
"\e[0m" (4 chars)

12+11+5+4 = 32

PS: "\e" is a GCC extension, use "\033" or "\x1b" for portability

does it not count the "" then??? thanks for the \033 tip

Originally Posted by cooper1200

does it not count the "" then??? thanks for the \033 tip

Ask yourself: How a string, in C, is defined and what is the purpose of strlen()?

bother it didnt print the \ that was supposed to be " \ " with no spaces sorry