When I compile this program, I keep getting this error
当我编译这个程序时,我一直得到这个错误。
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?
我认为这和指针有关。我该怎么解决这个问题。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';
}
1 个解决方案
#1
5
In your function h
you have declared that r
is a copy of a constant Record
-- therefore, you cannot change r
or any part of it -- it's constant.
在函数h中,你已经声明r是一个常数记录的拷贝——因此,你不能改变r或者它的任何一部分——它是常数。
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 copy of r
to the function h()
-- if you want to modify r
then you must pass a non-constant pointer.
还请注意,您正在向函数h()传递一个r的副本——如果您想修改r,那么您必须传递一个非常量指针。
void h( Record* r)
{
r->x *= 1000;
r->s[0] = 'C';
}
#1
5
In your function h
you have declared that r
is a copy of a constant Record
-- therefore, you cannot change r
or any part of it -- it's constant.
在函数h中,你已经声明r是一个常数记录的拷贝——因此,你不能改变r或者它的任何一部分——它是常数。
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 copy of r
to the function h()
-- if you want to modify r
then you must pass a non-constant pointer.
还请注意,您正在向函数h()传递一个r的副本——如果您想修改r,那么您必须传递一个非常量指针。
void h( Record* r)
{
r->x *= 1000;
r->s[0] = 'C';
}