assigning a char to a string

Question

I'm stuck here trying to understand why this assignement can't work in this way in C. What I'm trying to do is substitute all space occurrences with underscore char. (output: Hi_from_Synchronyze) I saw that the problem comes when I try to do this..

s[n]='_';

the complete code is this one

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

char *underscore(char *s, int n);

int main()
{
    printf("%s", underscore("Hi from Synchronyze", 0));
    return 0;
}

char *underscore(char *s, int n)
{
    if(s[n]=='\0')
        return s;
    else {
        if(s[n]==' ') {
            s[n]='_';
            return underscore(s, n+1);
        }
        else return underscore(s, n+1);
    }
}

I'd like to know what's going on behinde and why this happens, not the solution. Thank you very much in advance

`"Hi from Synchronyze"`. That is a constant string. It is readonly. It cannot be modified. — kaylum, Dec 06 '15 at 22:03
String literal can not be changed. change to `printf("%s", underscore((char[]){"Hi from Synchronyze"}, 0));` — BLUEPIXY, Dec 06 '15 at 22:03
It's [Compound Literal](http://en.cppreference.com/w/c/language/compound_literal). — BLUEPIXY, Dec 06 '15 at 22:08

score 2 · Answer 1 · answered Dec 06 '15 at 22:02

2

String literals are read-only, so you can't assign to them.

Make a mutable copy of the string first, something like:

char text[] = "Hi from Synchronyze";
printf("%s", underscore(text, 0));

answered Dec 06 '15 at 22:02

Emil Laine

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assigning a char to a string

1 Answers1