将字符串作为参数传递

Question

I am trying to pass string as an argument but result is always pointer.c:13:14: error: cannot convert 'char**' to 'char*' for argument '1' to 'int swg(char*)

the string i want to pass in looks like

char * str

this string gets value using fgets/getline

my function lookslike this

int swg ( char *g){
   char *tmp;
   size_t i;
   tmp=strtok(*g,">");
   for ( i = 0 ; i < strlen ( tmp ) ; i ++ ) {
     if(tmp[i]=='A') return 0;
   }
   return 1;
  }

and i am calling it like this

int tst;
tst=swg(str)

i even tried using tst=swg(&str) but it didnt work , how can i pass string as argument then?

Answer 1

The line that appears incorrect to me is

tmp=strtok(*g,">");

This should be:

tmp=strtok(g,">");

The call tst=swg(str) appears OK.

Answer 2

To simply fix the conversion mismatch, simply replace tmp=strtok(*g,">"); by tmp=strtok(g,">"); .

Anyways, you code appears to have other issues than this.

In order to process every token in your input string, you must call strtok() until there are no more tokens to be processed.

Find below an example of that.

#include <stdio.h>
#include <string.h>

int swg(char g[])
{
   size_t i;
   char *tmp = strtok(g, ">");
   while (tmp != NULL)
   {
       for (i=0; i<strlen(tmp); i++)
           if (tmp[i] == 'A')
               return 0;
       tmp = strtok(NULL, ">");
   }
   return 1;
}

int main()
{
    char str_one[] = "LOREM>A>IPSUM";
    char str_two[] = "LOREM>B>IPSUM";

    printf("Test one: %d\n", swg(str_one));
    printf("Test two: %d\n", swg(str_two));

    return 0;
}

Output:
Test one: 0
Test two: 1