passing in pointers as arguments to get string length

Question

I am playing with pointers in the K&R book and I wrote this program that swaps integers and measures the length of a string with a pointer. The first part works but my string length function does nothing. The program compiles and runs the first part and then the program stops responding.

    #include <stdio.h>

extern int a2 = 4;
extern int b2 = 5;

void swap(int *px, int *py);
int strlen2(char *s);
//int printLabel(char *thelabel, char newliner);

//int printLabel(char *thelabel, char newliner)
//{
//        int stringlength1=(strlen2(thelabel));
//  return stringlength1;
//}

void swap(int *px, int *py)  /* interchange *px and *py */
   {
       int temp;
       temp = *px;
       *px = *py;
       *py = temp;
   }

int strlen2(char *s)
   {
       int n;
       for (n = 0; *s != '\0', s++;)
           n++;
       return n;
   }

int main()
{
int a=4;
int b=5;
char newliner = '\n';
swap(&a,&b);
swap(&a2,&b2);
printf("%d",a);
printf("%c",newliner);
printf("%d",b);
printf("%c",newliner);
printf("%d",a2);
printf("%c",newliner);
printf("%d",b2);
printf("%c",newliner);
char sumstring[]="boo";
char *labelPtr;
labelPtr = sumstring;
int length = strlen2(labelPtr);
printf("%d",length);
return 0;
}

Also, try something like printf("%d\n", a); rather than the two printf lines. Also you could collapse statements like printf("%d\n%d\n%d\n", a, b, a2) and so on. — taronish4
– taronish4, Commented Aug 14, 2014 at 21:14
Minor: Better to use type size_t: size_t strlen2(char *s) { size_t n; ... and size_t length ... printf("%zu",length); — chux
– chux, Commented Aug 14, 2014 at 21:41

Chris Dodd · Accepted Answer · 2014-08-14 21:12:29Z

3

The problem is that this:

for (n = 0; *s != '\0', s++;)

is a semi-infinite loop. It checks for the terminating NUL, but then it ignores the result of that comparison and increements s, continuing the loop if it is non-null. Once it gets past the end of the string, the result is undefined behavior, but its likely to either loop forever or crash.

You probably meant

for (n = 0; *s != '\0'; s++)

answered Aug 14, 2014 at 21:12

Chris Dodd

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1 Comment

Amelia Over a year ago

Very strange. In the book, that's how it was shown. Thanks. That fixed it.

ppalancica · Accepted Answer · 2014-08-14 21:15:26Z

0

In the for loop, the second expression is usually a condition, in your case *s != '\0'. The 3rd expression is the increment, where you are supposed to increment the s pointer.

This is working fine:

int strlen2(char *s)
{
    int n;

    for (n = 0; *s != '\0'; s++)
        n++;

    return n;
}

answered Aug 14, 2014 at 21:15

ppalancica

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Comments

Jose Carcamo · Accepted Answer · 2014-08-14 21:19:15Z

0

Replace your code with the following:

int strlen2(char *s)
   {
       int n = 0;
       while(s[n] != '\0')
          ++n;
       return n;
   }

answered Aug 14, 2014 at 21:19

Jose Carcamo

1

Comments

David Ranieri · Accepted Answer · 2014-08-14 21:20:48Z

0

looks like a typo in the code. shouldn't this line be:

for (n = 0; *s != '\0'; s++)

instead of

for (n = 0; *s != '\0', s++;)

edited Aug 14, 2014 at 21:20

David Ranieri

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answered Aug 14, 2014 at 21:16

maruvatan

1

Comments

Vlad from Moscow · Accepted Answer · 2014-08-14 21:21:33Z

0

In this for statmenet

for (n = 0; *s != '\0', s++;)

in the condition part there is used the comma operator

*s != '\0', s++

Its results is the value of the last subexpression that is of s++. As pointer s is not equal to 0 then you get at least very long sycle.

I think you meant instead

for (n = 0; *s != '\0'; s++ )

answered Aug 14, 2014 at 21:21

Vlad from Moscow

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