在C中更改数组内部函数 [英] Changing array inside function in C

问题描述

我正在学习C并感到困惑，为什么在主函数中创建的数组不会在函数内部更改，我假设传递的数组是指针，并且更改指针应该更改了数组，对吗?有人可以解释这种情况下发生了什么吗?

I am learning C and confused why a array created in the main wont change inside the function, i am assuming the array passed is a pointer, and changing the pointer should've change the array , right ? can someone explain what is happening in this case?

寻求帮助.

int main(){
    int i, length = 10;
    int array[length];

    for (i = 0 ; i < length ; i++)
        array[i] = i * 10;
    printf("Before:");
    print(array, length);
    change(array, length);
    printf("After:");
    print(array, length);

    return 0;
}

// Print on console the array of int
void print(int *array,int length)
{
    int i;
    for(i = 0 ; i < length ; i++)
        printf("%d ", array[i]);
    printf("\n");
}

// Change the pointer of the array
void change(int *array,int length)
{
    int *new = (int *) malloc(length * sizeof(int));
    int i;
    for(i = 0 ; i < length ; i++)
        new[i] = 1;
    array = new;
}

我希望看到以下输出:

Before:0 10 20 30 40 50 60 70 80 90 
After:1 1 1 1 1 1 1 1 1 1 

我得到什么:

Before:0 10 20 30 40 50 60 70 80 90 
After:0 10 20 30 40 50 60 70 80 90 

推荐答案

在 c ，您不能通过引用传递变量，您在函数内部分配的array变量最初包含与传递的指针相同的地址，但它是其副本，因此对其进行修改不会更改传递的指针.

In c you can't pass a variable by reference, the array variable that you assign inside the function contains initially the same address as the passed pointer, but it's a copy of it so modifying it will not alter the passed pointer.

您需要像这样传递指针的地址，以便能够对其进行更改

You need to pass the address of the pointer in order to be able to alter it, like this

// Change the pointer of the array
void change(int **array, int length)
{
    *array = malloc(length * sizeof(int));
    if (*array == NULL)
        return;
    for (int i = 0 ; i < length ; i++)
        (*array)[i] = 1;
}

然后，在main()中，您无法分配给数组，通过这种函数执行此操作肯定是未定义的行为.在main()中定义的数组是在堆栈上分配的，您不能为该数组分配任何内容，因为它们是 non 可写的 lvalues ，因此您无法使其指向堆malloc()获得的内存位置，您需要传递这样的指针

Then in main() you cannot assign to an array, doing so through this kind of function is surely undefined behavior. The array defined in main() is allocated on the stack and you cannot assign anything to an array since they are non-writeable lvalues so you cannot make it point to a heap memory location obtained with malloc(), you need to pass a pointer like this

int *array;
change(&array, length);
free(array);

如果您希望函数替换之前的数组，则必须free()修改malloc()数据(请注意，将NULL传递给free()是明确定义的)，所以

If you want the function to replace the previous array, it will have to free() the malloc()ed data (note that passing NULL to free() is well defined), so

// Change the pointer of the array
void change(int **array, int length)
{
    free(*array);

    *array = malloc(length * sizeof(int));
    if (*array == NULL)
        return;
    for (int i = 0 ; i < length ; i++)
        (*array)[i] = 1;
}

然后在main()

int *array;
array = NULL;
change(&array, length);
change(&array, length);
change(&array, length);
change(&array, length);
free(array);

将执行您显然想要的操作.

will do what you apparently want.

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