C语言 错误：只读位置的分配
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error: assignment of read-only location
提问by monkey doodle
When I compile this program, I keep getting this error
当我编译这个程序时，我不断收到这个错误
example4.c: In function ‘h':
example4.c:36: error: assignment of read-only location
example4.c:37: error: assignment of read-only location
I think it has something to do with the pointer. how do i go about fixing this. does it have to do with constant pointers being pointed to constant pointers?
我认为这与指针有关。我该如何解决这个问题。它与指向常量指针的常量指针有关吗？
code
代码
#include <stdio.h>
#include <string.h>
#include "example4.h"
int main()
{
Record value , *ptr;
ptr = &value;
value.x = 1;
strcpy(value.s, "XYZ");
f(ptr);
printf("\nValue of x %d", ptr -> x);
printf("\nValue of s %s", ptr->s);
return 0;
}
void f(Record *r)
{
r->x *= 10;
(*r).s[0] = 'A';
}
void g(Record r)
{
r.x *= 100;
r.s[0] = 'B';
}
void h(const Record r)
{
r.x *= 1000;
r.s[0] = 'C';
}
回答by K Scott Piel
In your function hyou have declared that ris a copy of a constant Record-- therefore, you cannot change ror any part of it -- it's constant.
在您的函数中，h您已声明它r是常量的副本Record——因此，您不能更改r或更改它的任何部分——它是常量。
Apply the right-left rule in reading it.
在阅读时应用左右规则。
Note, too, that you are passing a copyof rto the function h()-- if you want to modify rthen you must pass a non-constant pointer.
也要注意，要传递一个副本的r的功能h()-如果要修改r，则必须通过非恒定的指针。
void h( Record* r)
{
r->x *= 1000;
r->s[0] = 'C';
}
