同一变量的不同内存地址

Question

Why is the address of two k different as shown in output of following code?

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
using namespace std;
int anu[1000000];
int calc(int a,int b,int c,int d)
{
    long long int k;
    k=(long long int)a*d*d+b*d+c;
    return k%1000000;
}
int main()
{
    int t,n,i,a,b,c,d,k;
    scanf("%d",&t);
    while(t--)
    {   
        scanf("%d %d %d %d %d",&n,&a,&b,&c,&d);
        memset(anu,0,sizeof(int)*1000000);
        anu[d]=1;
        vector<int> anu1;
        anu1.push_back(d);
        for(i=1;i<n;i++)
        {
            k=calc(a,b,c,anu1[i-1]);
            anu1.push_back(k);
            anu[k]=anu[k]?0:1;
        }
        d=0;k=0;
        printf("address of k=%d ",&k);
        for(i=0;i<1000000;i++)
        {
            if(anu[i])
                {
                if(d%2)
                k-=i;
                else
                k+=i;
                d++;
            }
        }
        printf("%d address of final k=%d\n",abs(k),&k);
    }
    return 0;
}

Input: 1 1 1 1 1 1

Output: Address of k=-1074414672 0 address of final k=1072693248

Answer 1

When I build with clang++ (with as many warnings enabled as I could) I get these warnings:

k.cpp:45:45: error: call to 'abs' is ambiguous
        printf("%d address of final k=%d\n",abs(k),&k);
                                            ^~~
/usr/local/include/c++/v1/cmath:660:1: note: candidate function
abs(float __x) _NOEXCEPT {return fabsf(__x);}
^
/usr/local/include/c++/v1/cmath:664:1: note: candidate function
abs(double __x) _NOEXCEPT {return fabs(__x);}
^
/usr/local/include/c++/v1/cmath:668:1: note: candidate function
abs(long double __x) _NOEXCEPT {return fabsl(__x);}
^

This is because you don't include <cstdlib> which declares the integer version of abs . Without that include the compiler have to guess which function it should use, and it seems it chooses poorly as it picks one of the floating point variants from <cmath> . This leads to you overwriting the next argument in printf call.

When building a program, I advise everyone to enable as many warnings as possible, they usually point out things like undefined behavior as in this case.

Answer 2

Without taking the address of the k variable, the compiler is allowed to use registers to hold the value of k .

With most processors, registers do not have a physical address. They are not in the CPU address space.

By printing the address of the k variable, you are telling the compiler to either store the variable in memory or for the compiler to generate an address.

Not all variables need to be stored in addressable memory; they can be store in registers.

Answer 3

1. You should print pointers with %p
2. abs() returns floating point number, casting it to int before printing solves the problem

The line would be correct with:

printf("%d address of final k=%d\n",(int)abs(k),&k);