指针数组

Question

请帮助修复此程序。 我尝试使用指针而不是数组打印指针数组但是我收到了这个错误：

pointer_multi_char4.c: In function ‘main’:
pointer_multi_char4.c:7:11: error: expected expression before ‘{’ token

这是代码：

#include <stdio.h>

int main (void){
char **message;
message= { "Four", "score", "and", "seven",
               "years", "ago,", "our", "forefathers" };
printf("%s\n",message);
return 0;
}

我该如何修复此代码？ 请有人解释一下该代码有什么问题

Answer 1

#include <stdio.h>

int main (){
    char *message[] =  { "Four", "score", "and", "seven",
                         "years", "ago,", "our", "forefathers", 0 };

    int loop;
    for (loop = 0; message[loop]; ++loop) printf("%s\n",message[loop]);
    return 0;
}

这种情况下的大括号（做出假设）是你想要初始化一个数组（因此使用char *message[]代替char **。

因为它是一个需要遍历它的数组。我使用空指针来标记数组的结尾

修改

然后@Lundu只需要

#include <stdio.h>

int main()
{
    const char * mesage="Four scor and seven years .... forefathers";
    printf("%s\n", message);
    return 0;
}

Answer 2

#include <stdio.h>

int main (void){
    char **message;
    message= (char* []){ "Four", "score", "and", "seven",
               "years", "ago,", "our", "forefathers" };
    int numOfMessage = 8;
    while(numOfMessage--){
        printf("%s\n", *message++);
    }
    return 0;
}