char*赋值为char [200]的不兼容类型

Question

执行main,它会要求输入.

将输入存储在argbuf中.

然后,使用strwrd将argbuf拆分为标记

然而,它说"错误:将char*分配给char [200]的不兼容类型"

我无法弄清楚为什么..

#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;
}

Answer 1

strwrd返回a char*,因此您无法将此值存储在char[200]变量中.请char*改用.