function tolower() not converting

Question

What is wrong with this piece of code? I can't find out what's going on.

#include <stdio.h>
#include <ctype.h>

int main(void)
{
    char *s = "OKAY";

    for (int i = 0; i < 4; i++)
        tolower(s[i]);

    printf("\n%s\n", s);

    return 0;
}

Output:

OKAY

Read the documentation of `tolower()`, and think about what your code does. — The Paramagnetic Croissant, May 28 '14 at 17:36
BTW, `tolower()` is so past-century. It doesn't handle some edge cases (german ß/SS, turkish I/ı/i, ...). You probably need something like ICU for proper case mapping. — ninjalj, May 28 '14 at 17:52

score 4 · Answer 1 · answered May 28 '14 at 17:37

4

The tolower function returns the lowercase equivalent of the input character. It doesn't modify it in place.

answered May 28 '14 at 17:37

Graeme Perrow

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haccks · Accepted Answer · 2014-05-28T17:51:32.923

4

You need to assign the return value of tolower to s but this will invoke undefined behavior because string literals are non-modifiable as they are placed on read only section of memory. You can't modify it. Try this instead

char s[]= "OKAY";
for (int i = 0; i < 4; i++)
    s[i] = tolower(s[i]);

edited May 28 '14 at 17:51

answered May 28 '14 at 17:38

haccks

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You are lucky that `"OKAY"` only contains characters from the basic character set. If it didn’t, the program would lead to *undefined behavior*. See any good documentation of `tolower` for the reason. – Roland Illig Jan 04 '15 at 22:10
@RolandIllig; Yeah, I know :) – haccks Jan 04 '15 at 22:31

score 1 · Answer 3 · answered May 28 '14 at 17:38

1

char s[] = "OKAY";

for (int i = 0; i < 4; i++)
    s[i]=tolower(s[i]);

answered May 28 '14 at 17:38

BLUEPIXY

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1

@haccks it provides no explanation and consequently it is unhelpful. – The Paramagnetic Croissant May 28 '14 at 17:42
1

@JoeFrambach I didn't flag this answer. – The Paramagnetic Croissant May 28 '14 at 19:46

function tolower() not converting

3 Answers3