错误:只读位置分配 [英] error: assignment of read-only location
问题描述
当我编译这个程序,我不断收到此错误
example4.c:在函数'H':
example4.c:36:错误:只读位置分配
example4.c:37:错误:只读位置分配
我认为这是与指针。我如何去修复这一点。它有恒定的指针做被指向到恒指针?
code
的#include<&stdio.h中GT;
#包括LT&;&string.h中GT;
#包括example4.h诠释的main()
{
记录值,* PTR; PTR =安培;价值; value.x = 1;
的strcpy(value.s,XYZ); F(PTR);
的printf(X%的D \\ n值,PTR - > X);
的printf(S的%s \\ n值,ptr-> S);
返回0;
}无效F(记录* R)
{
R-&X的催化剂* = 10;
(* R).S [0] ='A';
}无效克(记录r)
{
r.x * = 100;
R,S [0] ='B';
}无效小时(const的记录r)
{
r.x * = 1000;
R,S [0] =C;
}
在你的函数 ^ h
您已经声明研究
是一个常数录制
的副本 - 因此,你不能改变研究
或任何部分 - 这是不变的。
适用于阅读它的右 - 左的规则。
请注意,那就是,要传递一个复制 研究
到功能 H()
- 如果你想修改研究
,那么你必须通过一个非恒定的指针
无效小时(录音* R)
{
R-&X的催化剂* = 1000;
R-> S [0] =C;
}
When I compile this program, I keep getting this error
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?
code
#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';
}
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.
Apply the right-left rule in reading it.
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.
void h( Record* r)
{
r->x *= 1000;
r->s[0] = 'C';
}
这篇关于错误:只读位置分配的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!