Pass parameters in function

C c

创建时间:2019-02-22 23:57

字数:565 阅读:

pass by value Vs. pass by reference Vs. pass by pointer
1. pass by value

If you are not clear about the concept of pointer, please click here:

pass by value Vs. pass by reference Vs. pass by pointer

Firstly, in C language, there is only pass by value method to pass parameters. Therefore, we must use pointer to access the local variable of other function. pass by reference is the strategy supported by C++.

pass by value

void f( int  p){
    printf("\n%x",&p);            //12ff44
    printf("\n%x",p);            //10
    p=0xff;
}
void main()
{
    int a=0x10;
    printf("\n%x",&a);            //12fef4
    printf("\n%x\n",a);            //10
    f(a);
    printf("\n%x\n",a);            //10
}

In terms of the passing by value strategy, the variable will be copy to the memory as a new local variable(this is why the address of a is different than the p). Therefore, modifying the value of the p in the function f cannot modify the value of the a.

pass by reference

void f( int  &p){
    printf("\n%x",&p);            //12ff44
    printf("\n%x",p);            //10
    p=0xff;
}
void main()
{
    int a=0x10;
    printf("\n%x",&a);            //12ff44
    printf("\n%x\n",a);            //10
    f(a);
    printf("\n%x\n",a);            //ff
}

passing by reference strategy will pass the address of the actual parameter to the formal parameter, therefore, in fact, the variable p in the function c is same as the variable a. If we modify the value of p, the value of a also could be modified. This is why the address of both p and a are 12ff44, and the value of a has been modified to ff after calling the function f.

pass by pointer

void f( int  *p){
    printf("\n%x",&p);            //12fef4
    printf("\n%x",p);            //12ff44
    printf("\n%x",*p);            //10
    p=0xff;
}
void main()
{
    int a=0x10;
    printf("\n%x",&a);            //12ff44
    printf("\n%x\n",a);            //10
    f(a);
    printf("\n%x\n",a);            //ff
}

pass by pointer will also pass the address of the variable a to the pointer p in the function f. Note that 12fef4 is the actual address of the pointer p itself. And then the value of the pointer is 12ff44, i.e. the address of the a. Therefore the dereference operation will get the value of the variable a. And this method will also can change the value of a after calling the function f.

**Note: If you want to modify the value of a pointer, you still need to pass the address of this pointer to the other context(function).**

typedef struct ListNode
{
    char word[100];
    struct ListNode *next;
}node;

int main()
{
    ...
    node *prev;
    insert(root, cur, &prev); // must pass the address of the pointer
}

void insert(node *root, node *cur, node **prev)
{
    ...
    *prev = cur;
    // must pass the address of the pointer so as to modify the value of the original pointer
}

Here, if you only pass the prev instead of the &prev, it will be the pass by value. The pointer prev will be copied in the insert() scoop(stored in the stack segment, will be collected in the end of the function). Therefore, the actual value prev pointer has not been modified.

Reference

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