Question

所以我得到了一个我从教科书中输入的程序。

#include <stdio.h>
#define string char* 


int main(void) {
    string a[] = { "I", "like", "to", "fight," },
    b[] = {"pinch,","and","bite."};

    printf("%s %s %s %s %s %s %s\n", a[0], a[1], a[2], a[3], b[0], b[1], b[2]);

    return 0;
}

目前它会引发以下错误：

exit status 1
main.c: In function 'main':

main.c:7:18: error: excess elements in char array initializer
  b[] = {"pinch,","and","bite."};

main.c:7:18: note: (near initialization for 'b')

main.c:7:24: error: excess elements in char array initializer
  b[] = {"pinch,","and","bite."};

main.c:7:24: note: (near initialization for 'b')

显然，目标是让这项工作成功。该书暗示，在#define预处理器中，可以添加一个单个字符以使其工作。并且，由于错误都以b字符串为中心，我认为它是b。但事实并非如此。不管怎样，或者我把b置于错误的位置。我对这个问题的理解是，虽然它为字符串创建了足够的空间，但它不适用于b字符串。

任何意见都会受到赞赏。

Answer 1

好吧，如果你有

int *a, b;

你很快就会发现你有一个int指针和一个int。您的代码也会发生同样的事情

string a[] = { "I", "like", "to", "fight," },
b[] = {"pinch,","and","bite."};

变为

char* a[] = { "I", "like", "to", "fight," },
b[] = {"pinch,","and","bite."};

所以你有一个char指针数组和一个char数组。您需要添加的单个char是*，以使b成为一个char指针数组。

char* a[] = { "I", "like", "to", "fight," },
*b[] = {"pinch,","and","bite."};

C - 关于字符串的预处理器声明

1 个答案: