C - 错误:“a"的存储大小未知 [英] C - error: storage size of ‘a’ isn’t known

问题描述

这是我的 C 程序...

#include 结构 xyx {整数 x;输入 y;字符 c;字符 str[20];int arr[2];};int main(void){结构体 xyz a;a.x = 100;printf("%d\n", a.x);返回0；}

这是我得到的错误....

按 ENTER 或输入命令继续

<前>13structtest.c:在函数‘main’中:13structtest.c:13:13: 错误:a"的存储大小未知13structtest.c:13:13:警告:未使用的变量a"[-Wunused-variable]

解决方案

您的结构名为 struct xyx 但 a 的类型为 struct xyz>.修复后，输出为 100.

#include 结构 xyx {整数 x;输入 y;字符 c;字符 str[20];int arr[2];};int main(void){结构 xyx a;a.x = 100;printf("%d\n", a.x);返回0；}

This is my C program...

#include <stdio.h>

struct xyx {
    int x;
    int y;
    char c;
    char str[20];
    int arr[2];
};

int main(void)
{
    struct xyz a;
    a.x = 100;
    printf("%d\n", a.x);
    return 0;
}

This is the error that I am getting....

Press ENTER or type command to continue

13structtest.c: In function ‘main’:
13structtest.c:13:13: error: storage size of ‘a’ isn’t known
13structtest.c:13:13: warning: unused variable ‘a’ [-Wunused-variable]

解决方案

Your struct is called struct xyx but a is of type struct xyz. Once you fix that, the output is 100.

#include <stdio.h>

struct xyx {
    int x;
    int y;
    char c;
    char str[20];
    int arr[2];
};

int main(void)
{
    struct xyx a;
    a.x = 100;
    printf("%d\n", a.x);
    return 0;
}

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