C-使用'=='运算符，两个指针相等是什么意思？

Question

I'm a person in somewhere who is studying C language

I'm just curious of some parts of '==' operator.

I do know that pointer is a variable which stores address of memory. But here comes a question. When I tried to use '==' operator , eventhough those two pointers are pointing different address , the '==' operator worked.

Here's my code below

I want to know why the statement r == s considered to be True and why r and s gives me some 'non-sense value' which must be 90 ( in my computer those values are 56 )?

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 10
void add(int* , int* , int*);
int* Add(int* , int* , int*);
int main(){
    int a[N] , b[N] , c[N];
    int m ;
    int* p;
    int* q;
    int* r;
    int* s;
    char* ptr1 = "ATGC";
    char* ptr2 = "TCGA";
    char* ptr3 = ptr1;

    printf("result from strncmp : %d\n",strncmp(ptr1,ptr2,4));
    if(strncmp(ptr1,ptr2,4)==0){
        printf("strncmp operator works! those two are same\n");
    }else{
        printf("strncmp operator do not work! those two are not same\n");
    }
    printf("Okay another example here\n");
    if(ptr1 == ptr2){
        printf("ptr1 and ptr2 are same\n");
    }else if(ptr1==ptr3){
        printf("ptr1 and ptr3 are same");
    }else{
        printf("?\n");
    }
    printf("Now compare values after function call\n");
    for(m = 0 ; m < N ; m++){
        a[m] = m;
        b[m] = m*m;
        c[m] = 0;
    }
    printf("Works as void function\n");
    add(a,b,c);
    printf("After function call\n");
    for(m = 0 ; m < N ; m++){
        printf("%d + %d = %d\n",a[m],b[m],c[m]);
    }
    p = &c[N-1];
    q = &c[N-1];
    r = Add(a,b,c);
    s = Add(a,b,c);

    if(p==q){
        printf("Those two pointers p and q are same\n");
        printf("Address of pointer p is 0x%p\n",&p);
        printf("Address of pointer q is 0x%p\n",&q);
        printf("Value of pointer p is %d\n",*p);
        printf("Value of pointer q is %d\n",*q);
    }else{
        printf("Those are differenct\n");
    }

    if(r==s){
        printf("Those two pointers r and s are same\n");
        printf("Address of pointer r is 0x%p\n",&r);
        printf("Address of pointer s is 0x%p\n",&s);
        printf("Value of pointer r is %d\n",r);
        printf("Value of pointer s is %d\n",s);
    }else{
        printf("Those are differenct\n");
    }
    return 0;
}

void add(int* a , int* b , int* c){
    int i ;
    for(i = 0 ; i < N ; i++){
        c[i] = a[i] + b[i];
        printf("inside of  function loop : %d , current c is %d + %d = %d\n",i,a[i],b[i],c[i]);
    }
}

int* Add(int* a , int* b , int* c){
    int i ;
    for(i = 0 ; i < N ; i++){
        c[i] = a[i] + b[i];
        printf("inside of  function loop : %d , current c is %d + %d = %d\n",i,a[i],b[i],c[i]);
    }
}

Answer 1

You are not returning anything from Add function. But you are using its return value, this is undefined behavior.

You got r==s because both the times Add implicitly returned same arbitrary/garbage value.

Basically returning a value from function is same as popping out some value from the function's stack memory (and sometimes saving it in CPU Reg R0 ). When you do not specify which value to return, an arbitrary chosen value by compiler will be popped out and used.

To avoid such mistakes, always enable -Wall option (atleast) of your ( gcc ) compiler, and pay attention to every warning.

Answer 2

Because you printed a pointer, and address of this pointer but not the value it's pointing at.

Use *r and *s , if you want to print what it pointing at (holds).

printf("Those two pointers r and s are same\n");
printf("Address of pointer r is 0x%p\n",r);
printf("Address of pointer s is 0x%p\n",s);
printf("Value of pointer r is %d\n",*r);
printf("Value of pointer s is %d\n",*s);

or

printf("Value of pointer r is %d\n",r[0]);
printf("Value of pointer s is %d\n",s[0]);
printf("Value of pointer r is %d\n",r[1]);
printf("Value of pointer s is %d\n",s[1]);
printf("Value of pointer r is %d\n",r[2]);
printf("Value of pointer s is %d\n",s[2]);
printf("Value of pointer r is %d\n",r[3]);
printf("Value of pointer s is %d\n",s[3]); 

....

Answer 3

The function int* Add(int* a , int* b , int* c) is not returning any address to caller. The function Add can return global or heap memory to the caller. That is returning the stack(local variable) address to the caller still leads a various kind of problem.

Answer 4

You are calling the Add function, but you are not returning anything from the function. so it result undefined behavior.

Try the following change-

int* Add(int* a , int* b , int* c){
     int i ;
     for(i = 0 ; i < N ; i++){
         c[i] = a[i] + b[i];
         printf("inside of  function loop : %d , current c is %d + %d =    %d\n",i,a[i],b[i],c[i]);
     }
     return c; // Fix
}