无效的二进制操作数*(具有'ab {aka struct a}'和'ab * {aka struct a *}')

问题描述

我编写了一个程序来交换数组中的两个结构，我的编码如下

I wrote a program to swap two structures in an array and my coding is as follows

#include <stdio.h>
struct a {
    char *name;
    int id;
    char *department;
    int num;
};
typedef struct a ab;

void swap(ab *, ab *);

int main(int argc, char *argv[])
{
    ab array[2] = {{"Saud", 137, "Electronics", 500}, {"Ebad", 111, "Telecom", 570}};
    printf("First student data:\n%s\t%d\t%s\t%d", array[0].name, array[0].id,
           array[0].department, array[0].num);

    printf("\nSecond Student Data\n%s\t%d\t%s\t%d\n", array[1].name, array[1].id,
           array[1].department, array[1].num);

    swap(&array[0], &array[1]);
    // printf("")
    return 0;
}

void swap(ab *p, ab *q){
    ab tmp;
    tmp = *p
    *p = *q;
    *q = tmp;
}

编译时会出现错误，

怎么了?

推荐答案

第26行(上一行)的末尾缺少分号.

There's a missing semicolon at the end of line 26 (the previous line).

tmp=*p

由于这个原因，编译器认为下一行是同一条语句的一部分，这意味着整个语句将被解释为:

Due to this, the compiler considers the next line to be part of the same statement, meaning that the entire statement is interpreted as:

tmp=*p * p = *q;

第二个*被视为两个操作数的乘积-*p和p-这是错误消息来自的地方:

The second * is seen as a multiply of two operands - *p and p - which is where the error message comes from:

(因为*p是ab类型，而p是ab *类型).

(Because *p is of type ab, and p is of type ab *).

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