C将int数组指针作为参数传递给函数 [英] C pass int array pointer as parameter into a function

问题描述

我想将 B int 数组指针传递给 func 函数，并能够从那里更改它，然后查看 main 函数中的更改

I want to pass the B int array pointer into func function and be able to change it from there and then view the changes in main function

#include <stdio.h>

int func(int *B[10]){

}

int main(void){

    int *B[10];

    func(&B);

    return 0;
}

上面的代码给了我一些错误:

the above code gives me some errors:

In function 'main':|
warning: passing argument 1 of 'func' from incompatible pointer type [enabled by default]|
note: expected 'int **' but argument is of type 'int * (*)[10]'|

新代码:

#include <stdio.h>

int func(int *B){
    *B[0] = 5;
}

int main(void){

    int B[10] = {NULL};
    printf("b[0] = %d\n\n", B[0]);
    func(B);
    printf("b[0] = %d\n\n", B[0]);

    return 0;
}

现在我收到这些错误:

||In function 'func':|
|4|error: invalid type argument of unary '*' (have 'int')|
||In function 'main':|
|9|warning: initialization makes integer from pointer without a cast [enabled by default]|
|9|warning: (near initialization for 'B[0]') [enabled by default]|
||=== Build finished: 1 errors, 2 warnings ===|

推荐答案

在你的新代码中，

int func(int *B){
    *B[0] = 5;
}

B 是一个指向 int 的指针，因此 B[0] 是一个 int，你可以'不取消引用 int.只需删除 *,

B is a pointer to int, thus B[0] is an int, and you can't dereference an int. Just remove the *,

int func(int *B){
    B[0] = 5;
}

它有效.

在初始化中

int B[10] = {NULL};

您正在使用 void* (NULL) 初始化 int.由于存在从 void* 到 int 的有效转换，这是有效的，但它并不完全符合，因为转换是实现定义的，并且通常由程序员，因此编译器会对此发出警告.

you are initialising anint with a void* (NULL). Since there is a valid conversion from void* to int, that works, but it is not quite kosher, because the conversion is implementation defined, and usually indicates a mistake by the programmer, hence the compiler warns about it.

int B[10] = {0};

是 0 初始化 int[10] 的正确方法.

is the proper way to 0-initialise an int[10].

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