我在文件 test.c 中有一段简单的代码:
#include <stdio.h>
#include <string.h>
struct Student {
char name[10];
int grade;
};
int main(){
struct Student s = {.name = "Joe", .grade = 10}; //correct
struct Student students[10];
students[0] = {.name = "John", .grade = 10}; //error
students[1] = s; //correct
struct Student some_other_students[] = {
{.name = "John", .grade = 10}
}; //correct
students[2].grade = 8;
strcpy(students[2].name, "Billy"); //correct?!? O_o
return 0;
}
编译后出现此错误
test.c|14|error: expected expression before '{' token|
数组不应该在students[0] = {.name = "John", .grade = 10};
?正确初始化。以这种方式初始化时:struct Student s = {.name = "Joe", .grade = 10};
然后将其设置为这样的数组,students[1] = s;
我没有收到任何错误。只是一个假设,但是否可能是该{.name = "Joe", .grade = 10}
表达式在内存中没有引用,并且结构数组只是对已初始化结构的引用数组?
但话说回来,如果数组只是初始化结构的引用,这如何工作?students[2].grade = 8;
?
您只能在定义变量时对其进行初始化。各变量的定义后,当你不能使用相同的语法指定的变量,你使用的初始化。您尝试分配给数组元素无关紧要,重要的是您使用分配而不是初始化。
话虽如此,有一种方法可以解决这个问题,即使语法有点笨拙(IMO),那就是使用复合文字来创建一个临时的虚拟结构对象,并将其复制到变量中:
students[0] = (struct Student){.name = "John", .grade = 10};
以上等价于
{
// Define and initialize a "temporary" Student structure object
struct Student temp = {.name = "John", .grade = 10};
// Assign it to students[0]
students[0] = temp;
}
或者您可以在定义数组时对其进行初始化:
struct Student students[10] = {
{.name = "John", .grade = 10}
};
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