英文:
How to change the value of a variable passed as an argument and also use the value for an array
问题
这是我的示例代码：
#include <stdio.h>
void Func(int a[], int b) {
a[b] = 1;
b += 5;
}
int main(void) {
int a[10];
int b = 0;
printf("%d\n", b);
Func(a, b);
printf("%d\n", a[0]);
printf("%d\n", b);
}
我想让程序打印出：
0
1
5
我尝试通过将b改为指针来更改函数：
void Func(int a[], int *b)
... 并写上*b += 5。另外，我改变了函数的调用方式：
int a[10];
int *b = 0;
// ...
Func(a, &b);
能够操纵b并使用它吗？
注：我尝试遵循这个帖子。
英文:
This is my example code:
#include <stdio.h>
void Func(int a[], int b) {
a[b] = 1;
b += 5;
}
int main(void) {
int a[10];
int b = 0;
printf("%d\n", b);
Func(a, b);
printf("%d\n", a[0]);
printf("%d\n", b);
}
And I want the program to print:
0
1
5
I've tried changing the function by making b a pointer:
void Func(int a[], int *b)
... and by writing *b += 5. Also, I've changed the call of the function to:
int a[10];
int *b = 0;
// ...
Func(a, &b);
Is it possible to manipulate b and to use it?
Note: I've tried following this post
答案1
得分: 3
void Func(int a[], int *b) {
a[*b] = 1;
*b += 5;
}
int main(void) {
int a[10];
int b = 0;
printf("%d\n", b); // 打印变量 b 的值
Func(a, &b); // 将变量 b 的地址传递给 Func
printf("%d\n", a[0]);
printf("%d\n", b); // 再次打印变量 b 的值
}
Func 仍然有一个名为 b 的本地副本，但这个副本是一个指针，可以指向 main 中的 b 对象。这允许我们使用 *b += 5 修改指向的对象。
英文:
> c
> void Func(int a[], int b){
> a[b] = 1; // OK
> b += 5; // useless
> }
>
The problem with this code is that Func is assigning a local copy of b, and this has no effect outside of the function.
To get the expected output:
#include <stdio.h>
void Func(int a[], int *b) {
a[*b] = 1;
*b += 5;
}
int main(void) {
int a[10];
int b = 0;
printf("%d\n", b); // print value of b
Func(a, &b); // pass address of b to Func
printf("%d\n", a[0]);
printf("%d\n", b); // print value of b again
}
Func still has a local copy named b, but this copy is a pointer which may point to the b object in main. This allows us to modify the pointed-to object with *b += 5.
答案2
得分: 1
你需要将a和b用作指针，因为a是一个数组，而你想要在函数内修改b的值。
但是你不能直接将b直接传递给函数，这就是为什么要将其作为&b发送，以提供变量存储的地址，这样你就不会复制b的值，而是直接改变它。
void foo(int *a, int *b) {
a[*b] = 1;
*b += 5;
}
int main(void) {
int a[10] = {0};
int b = 0;
printf("b = [%i]\n", b);
foo(a, &b);
printf("a[0] = [%i]\n", a[b]);
printf("b = [%i]\n", b);
}
英文:
For that you want to use a and b as a pointer, because a is an array and b you want to modify the value inside the function.
But you can't send directly b inside the function is why you to send as &b to give the address where the variable is stored in that way you don't copy the value of b but directly change it.
void foo(int *a, int *b) {
a[*b] = 1;
*b += 5;
}
int main(void) {
int a[10] = {0};
int b = 0;
printf("b = [%i]\n", b);
foo(a, &b);
printf("a[0] = [%i]\n", a[b]);
printf("b = [%i]\n", b);
}
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