Make word lowercase in C with function

Question

I'm not sure what the problem is. The function seems to be working well, but in the main function, when testing with a printf, it doesn't show the result.

char* MotMiniscule(char* mot)
{
char motm[100],c,*motf;
int i=0;
strcpy(motm,mot);
c=motm[i];
while(c!='\0')
{
    printf("%c \n",c);
    motm[i]=tolower(c);
    i++;
    c=motm[i];
}
strcpy(motf,motm);
printf("%s\n",motf);
return(motf);
}

main()
{
char *mot="HEllooOoMA",*min;
min=MotMiniscule(mot);
printf("\n le mot est %s:\n",mot);
printf("|| %s  ||",min);
}

`main()` returns `int`. Does your book or tutorial say otherwise? If so, replace it. Omitting the return type is a syntax error and makes your code ill-formed. — underscore_d, Dec 21 '17 at 13:51

iBug · Answer 1 · 2017-12-21T14:06:26.513

2

You've never allocated space for the pointer motf in function MotMiniscule:

strcpy(motf,motm);

This is undefined behavior because the address in motf is indeterminate. You should give it some space to point to:

motf = malloc(100);

The full code should be:

char* MotMiniscule(char* mot)
{
    char motm[100],c,*motf;
    int i=0;
    strcpy(motm,mot);
    c=motm[i];
    while(c!='\0')
    {
        printf("%c \n",c);
        motm[i]=tolower(c);
        i++;
        c=motm[i];
    }
    motf = malloc(100); // Allocate some memory
    strcpy(motf,motm);
    printf("%s\n",motf);
    return(motf);
}

int main()
{
    char *mot="HEllooOoMA",*min;
    min=MotMiniscule(mot);
    printf("\n le mot est %s:\n",mot);
    printf("|| %s  ||",min);
    free(min); // Don't forget to free dynamically allocated memory
}

As pointed out by John Bode, the use of motm is totally redundant. You can safely remove it. Besides, the size of the dymanical allocation should rely on the length of mod. So a refined version of the code is this.

char* MotMiniscule(char* mot)
{
    char c, *motf;
    int i = 0;
    c = mot[0];
    motf = malloc(strlen(mot) + 1); // Allocate some memory
    while (c != '\0')
    {
        printf("%c\n", c);
        motf[i] = tolower(c);
        i++;
        c = mot[i];
    }
    // No need to copy again, but
    motf[i] = '\0'; // Remember to terminate it
    printf("%s\n", motf);
    return(motf);
}

int main()
{
    char *mot = "HEllooOoMA", *min;
    min = MotMiniscule(mot);
    printf("\n le mot est %s:\n", mot);
    printf("|| %s  ||", min);
    free(min); // Remember to free it
}

edited Dec 21 '17 at 14:06

answered Dec 21 '17 at 13:51

iBug

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And call `free(min);` at the end of `main()` to deallocate the memory. The program ends at that point and on any modern platform will be recovered in tear-down of the process but it's just bad practice to not match `malloc()` with `free()`. – Persixty Dec 21 '17 at 13:55
1

Quibble - the argument to `malloc` should be based on the size of the input string, not just some random constant. – John Bode Dec 21 '17 at 13:56
1

@Persixty I added it, right...? – iBug Dec 21 '17 at 13:56
@JohnBode The OP wrote `char motm[100]` at the beginning of the function. – iBug Dec 21 '17 at 13:56
@iBug Yup. You did. :) – Persixty Dec 21 '17 at 13:57
@iBug: which is just as wrong. The size of `motf` should be `strlen( mot ) + 1`, and `motm` should be eliminated entirely - it’s redundant. – John Bode Dec 21 '17 at 14:00
@JohnBode Useful comment, thanks. But that'd require extra changes to the code. I'll explain it in the upcoming paragraphs. – iBug Dec 21 '17 at 14:02
@JohnBode How does that look? – iBug Dec 21 '17 at 14:06
@iBug: Perfect. – John Bode Dec 21 '17 at 14:13

Make word lowercase in C with function

1 Answers1