incompatible types in assignment of char * to char[200]

Question

Execute main, it will ask for input.

Store input in the argbuf.

Then, use strwrd to split argbuf into tokens

Yet, it says "Error: incompatible types in assignment of char * to char[200]"

I cannot figure out why..

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char argbuf[200];//used to store input

char *strwrd(char *s, char *buf, size_t len, char *delim){
    s += strcspn(s, delim);
    int n = strcspn(s, delim); /* count the span (spn) of bytes in */
    if (len-1 < n)             /* the complement (c) of *delim */
        n = len-1;
    memcpy(buf, s, n);
    buf[n] = 0;
    s += n;
    return (*s == 0) ? NULL : s;
}



int main(){
    fgets(argbuf, sizeof(argbuf), stdin);
    char token[10][20];
    int index;
    for (index = 0; index < 10; index++) {
        argbuf = strwrd(argbuf, token[index], sizeof(token[index]), " \t");
        if (argbuf == NULL)
            break;
    }
    return 1;
}

Thanks for reminding me that. I always try to answer questions...but, i am new to programming... — mabeiyi, Sep 17 '12 at 17:48
possible duplicate of [Difference between char* and char\[\] in C](http://stackoverflow.com/questions/10054271/difference-between-char-and-char-in-c) — netcoder, Sep 17 '12 at 18:02

score 5 · Accepted Answer · answered Sep 17 '12 at 16:56

5

strwrd returns a char*, so you can't store this value in a char[200] variable. Use char* type instead.

answered Sep 17 '12 at 16:56

md5

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I think "argbuf" represents the address of argbuf[200]? – mabeiyi Sep 17 '12 at 16:57
You've declared argbuf as an array, so at times it's appropriate to treat it like a char*. But here, you can't return a char* into argbuf because it's immobile in memory. – desimusxvii Sep 17 '12 at 17:00
I declared another pointer "char *buf=NULL". and assign the address of the array to buf "buf=argbuf". Then, use the buf as the parameter of strwrd. Finally, it works. – mabeiyi Sep 17 '12 at 17:45

score 1 · Answer 2 · edited May 23 '17 at 11:51

1

char * and char[x] are different types, see here

Here's another good resource

In your code char argbuf[200]; is a statically allocated array, so you can not assign a pointer to it. Why pass the global back and forth anyway? If you're going to use argbuf as a global variable then just modify it inside strwrd directly if your each for '\t' comes back with something valid.

edited May 23 '17 at 11:51

Community

1
1

answered Sep 17 '12 at 17:23

Mike

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incompatible types in assignment of char * to char[200]

2 Answers2