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我寫了一個程序來交換數組中的兩個結構,我的編碼如下無效的二進制操作數*(有'ab {aka struct a}'和'ab * {aka struct a *}')
#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;
}
在編譯它,它給出了一個錯誤,
newproject.c: In function ‘swap’:
newproject.c:26:3: error: invalid operands to binary * (have ‘ab {aka
struct a}’ and ‘ab * {aka struct a *}’)
*p=*q;
什麼是錯?
有時錯誤不在報告的行上,而是在以前的行中。看看前面的一行,看看看起來好不好。例如,也許不會錯過分號? –
[爲什麼編譯器報告缺少分號?](http://stackoverflow.com/questions/40135392/why-doesnt-compiler-report-missing-semicolon) –