Code works fine if the name of the string is given as parameter in function but on giving the string itself as parameter it generates error

Question

#include<stdio.h>
int main(void)
{
    char str1[10]="Versace";
    int length(char *p);
    char* reverse(char *p);

    printf("Length of the array is %d\n",length("Versace")); 

    printf("Reversed string %s",reverse("Versace"));
}
int length(char*p)
{
     int i;
     for(i=0;*(p+i)!='\0';i++);
     return i;
}
char* reverse(char *p)
{
     int i,l;
     char temp;
     for(l=0;*(p+l)!='\0';l++);
     
     for (i=0;i<(l/2);i++)
     {
           temp=*(p+i);
           *(p+i)=*(p+l-1-i);
           *(p+l-1-i)=temp;
    }
     return(p);

}

I have written a code to find the length of a string and to reverse a string.For this purpose I've made 2 functions reverse and length. But, why does this code show error? The code works fine if I use str1 as parameter in reverse function in the second printf.

Is this homework and you can't use standard library functions? — Schwern, Oct 28 '20 at 05:00
Instead of `*(p+i)`, `*(p+l-1-i)` and such, just write `p[i]`, `p[l - 1 - i]` — Antti Haapala -- Слава Україні, Oct 28 '20 at 06:16

Schwern · Answer 1 · 2020-10-28T06:37:49.197

1

The compiler warning -Wwrite-strings offers a clue.

test.c:9:49: warning: passing 'const char [8]' to parameter of type 'char *' discards qualifiers
      [-Wincompatible-pointer-types-discards-qualifiers]
    printf("Length of the array is %d\n",length("Versace"));

"Versace" is stored in the executable itself which is mapped to appear to be read-only memory. If you look at your executable in an editor, you will see Versace among the noise. It is read-only and you're trying to modify it.

If we make the change the warning suggests and have reverse take a const char * we get a compiler error because you can't change a constant.

test.c:22:16: error: read-only variable is not assignable
         *(p+i)=*(p+l-1-i);
         ~~~~~~^
test.c:23:20: error: read-only variable is not assignable
         *(p+l-1-i)=temp;

char str1[]="Versace"; also stores "Versace" in the executable, but it then copies it into real memory which can be modified.

edited Oct 28 '20 at 06:37

answered Oct 28 '20 at 05:18

Schwern

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Unfortunately C string literals are arrays of *unqualified* chars... – Antti Haapala -- Слава Україні Oct 28 '20 at 05:58
Relevant in that there are no compiler warnings offering a clue, if your compiler conforms to standard C. – Antti Haapala -- Слава Україні Oct 28 '20 at 06:13
And what compiler you are using? And there is no requirement that "Versace" is read-only, only that modifications will have undefined behaviour. – Antti Haapala -- Слава Україні Oct 28 '20 at 06:23
and how do you produce the warning because neither my GCC 9.3.0 nor clang 10.0.0 does produce any diagnostics on that line. – Antti Haapala -- Слава Україні Oct 28 '20 at 06:25
Well, your answer did not tell about the use of `-Wwrite-strings` (which is actually non-conforming, because it changes the type of string literals). – Antti Haapala -- Слава Україні Oct 28 '20 at 06:31

Code works fine if the name of the string is given as parameter in function but on giving the string itself as parameter it generates error

1 Answers1