search of string in an array of strings - c

Question

i wrote some code that is supposed to find the location of a given string in an array of strings.
problem is- it doesn't give the location. it gives something else.
i understand that probably the problem has to do with the differences between the pointers that are involved- a previous version that dealt with finding the position of a letter in a word worked well.
after a lot of attempts to figure out where is the bug, i ask your help.
kindly, explain me what should be done.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int what (char * token);
main()
{
int i=0;
char string[]="jsr";
char *token;
token=&string[0];
i=what(token);
printf(" location of input is %d \n", i);
return 0;
}
int what (char * token)
{
int i=1;
char *typtbl[]={"mov",
"cmp",
"add",
"sub",
"not",
"clr",
"lea",
};
char * ptr;
ptr=(char *)typtbl;
while (!(strcmp(ptr,token)==0))
{
ptr=(char *)(typtbl+i);
i++;
}
return i;
}

Answer 1

As pointed out, you did not design function what properly. What value should it return if your search function go through all the pointers but does not find the desired string? Typically in that case return -1 would be a choice to indicate nothing found. Also in this case, using a for loop would probably be more suitable, you can just return the index immediately instead of going through all pointers.
int what(char *token)
{
char *typtbl[] = {
"mov",
"cmp",
"add",
"sub",
"not",
"clr",
"lea",
};
for( size_t i = 0; i < sizeof(typtbl)/sizeof(char*); ++i )
{
char *ptr = typtbl[i];
if(strcmp(ptr, token) == 0)
{
return i; // found something
}
}
return -1; // found nothing
}

Answer 2

A cleaner working version.
Main issue is in the (char *)(typtbl+i) replaced by typtbl[i] in the following code. typtbl+i is equivalent to &typtbl[i], so if my memory is good, it's a pointer on the pointer of the string and not the pointer of string itself
I added a NULL at the end of the array to be able to stop if the string is not present and return -1 to clearly say it was not found.
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
int what(char *token);
int main()
{
int i = 0;
char string[] = "jsr";
i = what(string);
printf(" location of input is %d \n", i);
return 0;
}
int what(char *token)
{
char *typtbl[] = {
"mov",
"cmp",
"add",
"jsr",
"not",
"clr",
"lea",
NULL
};
int i = 0;
while(typtbl[i] && !(strcmp(typtbl[i], token) == 0)) {
++i;
}
if(!typtbl[i])
i = -1;
return i;
}
char *token; token=&string[0]; was useless because string == &string[0].

Answer 3

A few things:
Your main function is missing its return type.
The while loop in what doesn't stop when the element isn't found. Therefore you are reading out of bounds.
This should do the work w/o pointer arithmetic.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int what (char * token);
int main(){
int i=0;
char string[]="jsr";
char *token;
token=&string[0];
i=what(token);
printf(" location of input is %d \n", i);
return 0;
}
int what (char * token){
unsigned int i=0;
char *typtbl[]={"mov",
"cmp",
"add",
"sub",
"not",
"clr",
"lea",
};
unsigned int typtbl_x_size = sizeof(typtbl)/sizeof(typtbl[0]);
char * ptr;
ptr=typtbl[i];
while (!(strcmp(ptr,token)==0)){
i += 1;
if (i >= typtbl_x_size){
printf("element not in list\n");
return -1;
}
ptr=typtbl[i];
}
return i;
}