While loop error while implementing 'strend' - c

Question

I keep getting errors when I run the program I wrote for 'strlen'. (strlen returns 1 when one string appears at the end of another). I guess it's the first while loop that gives the error, because when I put printf("%c\n", *s); instead of empty ;, it just works fine. Is there something wrong with the syntax???
#include <stdio.h>
int strend(char *s, char *t)
{
int len;
int dummy;
while ( *s++ )
; // why error???
while ( *t++ )
len++;
for ( ; len>0 ; len--)
{
if ( *(s-len) != *(t-len) )
return 0;
}
return 1;
}
int main() {
char one[] = "I 0dont like youa";
char two[] = "ke youa";
printf("%d\n", strend(one, two));
}

Answer 1

First of all you forgot to initialize len, you should initialize it with 0.
int len=0;
Secondly use this:
while ( *s )
s++; // s should be incremented only if *s is not 0.
while ( *t )
{
t++; //Same issue here.
len++;
}

Answer 2

First of all local variable len was not initialized
int len;
So the function already has undefined behaviour.
Second after the loops
while ( *s++ );
while ( *t++ ) len++;
the both pointers point to beyond the terminating zeroes of each strings. As result this comparison (along with the for loop itself)
if ( *(s-len) != *(t-len) )
is not valid.
Indeed let's assume that we have
char s[] = { 'a', '\0' };
char t[] = { 'a', '\0' };
After the above loops len will be equal to 1 (provided that initially you initialized len with 0 ). At the same time t and s will point to the memory after the terminating zero. So *( s - len ) and *( t - len ) will represent elements '\0' and you will compare only these terminating zeroes while the elements with 'a' will not be compared because the condition of the for loop is len>0. That means that there will be only one iteration of the loop when len is equal to 1.
A correct function can be written simpler and more clear
int strend( const char *s, const char *t )
{
const char *p = s;
const char *q = t;
while ( *s++ );
while ( *t++ );
while ( s != p && t != q && *( s - 1 ) == *( t - 1 ) ) --s, --t;
return t == q;
}
If it is unimportant which one string appears at the end of another string then simply substitute this return statement
return t == q;
for the following one
return s == p || t == q;
In this case this program
//...
int main( void ) {
char one[] = "I 0dont like youa";
char two[] = "ke youa";
printf("%d\n", strend(one, two));
printf("%d\n", strend(two, one));
}
will output
1
1
While if to use use the first return statement return t == q; then the program
//...
int main( void ) {
char one[] = "I 0dont like youa";
char two[] = "ke youa";
printf("%d\n", strend(one, two));
printf("%d\n", strend(two, one));
}
will output
1
0

Answer 3

The error should come because of the missing initialization int len = 0;. I recommend using while always with embraces like this while { /*your code*/ };.
Your 1st while loop should be fine anyway. I made some modifications with two separated loops.
#include <stdio.h>
int strend(char *s, char *t)
{
int len = 0; // init 0
int i = 0,
j = 0;
while (*s++) { i++; } // get both string lengths
while (*t++) { j++; }
if (i > j) // compare and use shorter one
len = j;
else
len = i;
len++; // offset: start from 1st character
// [fix-issue: 2015-06-28 12:41]
for ( ; len>0 ; len--)
{
//printf("%d: \"%c\" \"%c\"\n", len, *(s-len), *(t-len)); // test-out
if ( *(s-len) != *(t-len) )
return 0;
}
return 1;
}
int main()
{
char one[] = "ABCDE";
char two[] = "CDE";
printf("test case 1: [%s][%s] %d\n", one, two, strend(one, two));
two[1] = '_';
printf("test case 2: [%s][%s] %d\n", one, two, strend(one, two));
two[0] = '?';
two[1] = 'D';
printf("test case 3: [%s][%s] %d\n", one, two, strend(one, two));
two[0] = 'C';
two[2] = '#';
printf("test case 4: [%s][%s] %d\n", one, two, strend(one, two));
}
The output goes like this:
test case 1: [ABCDE][CDE] 1
test case 2: [ABCDE][C_E] 0
test case 3: [ABCDE][?DE] 0
test case 4: [ABCDE][CD#] 0