[英]How can I change an int array in a function
Im new with c language and im having difficult while trying to change an array on my main function using another function. 我是C语言的新手,尝试使用其他函数更改我的主要函数的数组时遇到困难。 Besides seeing the solution I'd also like to get a full explanation what is wrong with my code and what's the explanation to your solution.
除了查看解决方案之外,我还想获得完整的解释,我的代码有什么问题以及您的解决方案的解释是什么。
OK so I did too much tryouts and error experimentation but with no success to solve my problem. 好的,所以我进行了过多的测试和错误实验,但未能成功解决问题。 Eventually this is my current code:
最终这是我当前的代码:
#include <stdio.h>
#define MAX 20
typedef int Values[MAX];
int changeArr(Values *vals2)
{
*vals2[0] = 200;
*vals2[1] = 100;
printf("%d and ", *vals2[0]);
printf("%d\n", *vals2[1]);
return 0;
}
int main (int argc, char *argv[])
{
Values vals;
changeArr(&vals);
printf("%d and ", vals[0]);
printf("%d\n", vals[1]);
return 0;
}
Output is: 输出为:
200 and 100
200 and 0
Instead of: 代替:
200 and 100
200 and 100
Version 1: 版本1:
#include <stdio.h>
#define MAX 20
typedef int Values[MAX];
int changeArr(int vals2[]) {
vals2[0] = 200;
vals2[1] = 100;
printf("%d and ", vals2[0]);
printf("%d\n", vals2[1]);
return 0;
}
int main (int argc, char *argv[]) {
Values vals;
changeArr(vals);
printf("%d and ", vals[0]);
printf("%d\n", vals[1]);
return 0;
}
Instead of passing a pointer to the array, pass a pointer to the array's first element. 而不是将指针传递给数组,而是将指针传递给数组的第一个元素。
Version 2: 版本2:
#include <stdio.h>
#define MAX 20
typedef int Values[MAX];
int changeArr(Values *vals2) {
(*vals2)[0] = 200;
(*vals2)[1] = 100;
printf("%d and ", (*vals2)[0]);
printf("%d\n", (*vals2)[1]);
return 0;
}
int main (int argc, char *argv[]) {
Values vals;
changeArr(&vals);
printf("%d and ", vals[0]);
printf("%d\n", vals[1]);
return 0;
}
use parentheses to compensate for the precedence. 使用括号来补偿优先级。
The problem in your code is that 您的代码中的问题是
*vals2[1]
is *(vals2[1])
, so it dereferences a pointer one unit of Values
after the passed pointer. 是
*(vals2[1])
,因此它在传递的指针之后以Values
一个单位解引用指针。 You have no accessible memory allocated there, so it's undefined behaviour. 您那里没有分配可访问的内存,因此它是未定义的行为。 In practice, it is the same as accessing
实际上,它与访问
arr[1][0]
for an 为
int arr[2][MAX];
But your vals
is only (equivalent to) an int arr[1][MAX];
但是您的
vals
只(相当于)的int arr[1][MAX];
, so you are accessing out of bounds. ,因此您无法访问。 But if nothing worse happens,
*vals2[1] = 100;
但是如果没有更糟的情况发生,
*vals2[1] = 100;
in changeArr
sets vals[MAX]
in main
to 100. It may overwrite something crucial, though. 在
changeArr
将main
vals[MAX]
设置为100。但是,它可能会覆盖一些至关重要的内容。
In int changeArr(Values *vals2)
you are passing a pointer to an array of MAX
int
s, resp the first such array in an array of Values
. 在
int changeArr(Values *vals2)
您传递了一个指向MAX
int
数组的指针,分别是Values
数组中的第一个此类数组。 Then vals2[1]
is the second array in that array of Values
(which doesn't exist here), and *vals2[1] == vals2[1][0]
the first int
in that second array. 然后
vals2[1]
是该Values
数组中的第二个数组(此处不存在), *vals2[1] == vals2[1][0]
是第二个数组中的第一个int
。 You want to modify elements of the first (and only) array in the pointed-to memory block, so you want to access vals2[0][1]
, or equivalently (*vals2)[1]
. 您想要修改指向的内存块中第一个(也是唯一一个)数组的元素,因此要访问
vals2[0][1]
或等效地(*vals2)[1]
。
A picture: 照片:
vals2 vals2[1]
| |
v v
|vals[0]|vals[1]|...|vals[MAX-1]|x
in changeArr
, the pointer vals2
points to the array vals
. 在
changeArr
,指针vals2
指向数组vals
。 Since it's a pointer to an int[MAX]
, vals2+1
points to an int[MAX]
at an offset of MAX*sizeof(int)
bytes after the start of vals
(which is just behind the end of vals
). 由于它是一个指向
int[MAX]
vals2+1
指向的int[MAX]
在一个偏移MAX*sizeof(int)
字节开始后vals
(它是仅次于的端vals
)。 vals2[1]
, or equivalently *(vals2 + 1)
, is that array just after vals
(which doesn't exist). vals2[1]
或等效的*(vals2 + 1)
是vals
之后的那个数组(不存在)。
You want to change vals[1]
, which is located in the array vals2[0]
, at an offset of 1*sizeof(int)
bytes, so you need vals2[0][1]
or equivalently (*vals2)[1]
. 您想要更改
vals[1]
(位于数组vals2[0]
,其偏移量为1*sizeof(int)
个字节,因此您需要vals2[0][1]
或等效地(*vals2)[1]
。
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