为什么这段代码不起作用？ （指针） [英] why this code does not work? (pointers)

问题描述

#include<stdio.h>
#include<string.h>

void allocate(int *arr,int size,int value){
arr = (int *) malloc(size*sizeof(int));
int i =0;
while(i<size)
{
    *(arr+i) = value;
    i++;
}
}
int main()
{

 int *vector = NULL;
 allocate(vector,5,45);
 printf("%d",*vector);

}

推荐答案

因为C按值传递所有变量，而不是通过引用传递。

这意味着当您将值传递给方法时，您在方法内所做的更改不会影响调用它的函数中的变量：

Because C passes all variables by value, not by reference.
What that means is that when you pass a value to a method, changes you make inside the method do not affect the variable in the function that called it:

void doit(int i)
   {
   i++;
   }
...
   i = 3;
   doit(i);
   printf("%d", i);
...

将打印3，而不是4。

这是明智的：否则我可以这样做：

Will print "3", not "4".
This is sensible: otherwise I could do this:

doit(3);
printf("%d", 3);

然后事情会变得非常混乱。

所以当你传递指向你的分配函数的指针时，它会将变量中的值（NULL）传递给函数，而不是对保存null的位置的引用。

有两种方法可以解决这个问题：

1）更改分配以接受指向a的指针-pointer-to-an-integet，向量的dpass地址：

And then things would get very confusing.
So when you pass a a pointer to your allocate function, it passes the value in the variable (NULL) to the function, and not a reference to the location that hold the null.
There are two ways to solve this:
1) change allocate to accept a pointer-to-a-pointer-to-an-integet, an dpass teh address of vector:

void allocate(int **arr,int size,int value){
*arr = (int *) malloc(size*sizeof(int));

...

 int *vector = NULL;
 allocate(&vector,5,45);

...

2）获取分配以返回值：

2) Get allocate to return the value instead:

int* allocate(int size,int value){
int *arr = (int *) malloc(size*sizeof(int));
int i =0;
while(i<size)
{
    *(arr+i) = value;
    i++;
}
return arr;
}
int main()
{

 int *vector = NULL;
 vector = allocate(5,45);

我？我将返回第二个版本中的值。

Me? I'd return the value as in the second version.

您的代码正在修改allocate函数中的局部变量，该函数不会返回给main。您需要传递指针的地址，如下所示：

Your code is modifiying the local variable in the allocate function, which is not returned to main. You need to pass the address of the pointer as below:

void allocate(int **arr, int size, int value)
{
    *arr = (int *)malloc(size * sizeof(int));
    int i = 0;
    while(i < size)
    {
        *(*arr + i) = value;
        i++;
    }
}
int main()
{
    int *vector = NULL;
    allocate(&vector, 5, 45);
    printf("%d",*vector);
}

虽然更好（更具可读性）的版本是分配函数将指针返回到main，例如：

Although a better (and more readable) version would be for the allocate function to return the pointer to main, like:

int* allocate(int size, int value)
{
    int* arr = (int *)malloc(size * sizeof(int));
    int i = 0;
    while(i < size)
    {
        *(arr + i) = value;
        i++;
    }
    return arr;  // thanks to Jochen for reminding me about this
}
int main()
{
    int *vector = allocate(5, 45);
    printf("%d",*vector);
}

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