当尝试将新的字符串值分配给 c 中现有的 char 数组时,出现以下错误:
assignment to expression with array type
employees[id].FirstMiddleName = NewFirstMiddleName;
^
我认为这两个变量都是相同大小的数组,所以我不明白错误指的是什么或如何修复它。
struct employee {
char LastName[30];
char FirstMiddleName[35];
float Salary;
int YearHired;
};
int modify(int id) {
char NewLastName[30];
char NewFirstMiddleName[35];
float NewSalary;
int NewYearHired;
printf("Enter new first (and middle) name(s): \n");
gets(NewFirstMiddleName);
employees[id].FirstMiddleName = NewFirstMiddleName;
printf("Enter new last name: \n");
gets(NewLastName);
employees[id].LastName = NewLastName;
....
}
int main() {
struct employee *ptr, person;
ptr = &person;
ptr->LastName[0] = '\0';
ptr->FirstMiddleName[0] = '\0';
ptr->Salary = -1;
ptr->YearHired = -1;
for(int i = 0; i < 20; i++) {
employees[i] = person;
//printf("%i\n", i);
}
....
}
最佳答案
employees[id].FirstMiddleName
是数组类型,不能使用 =
assignment operator shall have a modifiable lvalue as its left operand. A modifiable lvalue is an lvalue that does not have array type
在这种情况下,您需要使用strcpy
strcpy(employees[id].FirstMiddleName, NewFirstMiddleName);
关于c - 将新字符串值分配给 char 数组时出错,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/53975030/