在scanf（）读取结果导致进程终止后，C ++删除指针

Question

Could you explain to me why on earth this code terminates the process?

#include<cstdio>

int main(){
    int *wsk = new int[2];

    scanf("%d %d", wsk[0], wsk[1]); // relevant line

    delete [] wsk;
    wsk = new int[10];
    return 0;
}

Whereas this one does not:

#include<cstdio>

int main(){
    int *wsk = new int[2];

    scanf("%d %d", &wsk[0], &wsk[1]); // relevant line, mark the two '&'

    delete [] wsk;
    wsk = new int[10];
    return 0;
}

Answer 1

scanf takes pointers.

It has to know the address of where to put the read in value.

wsk[0] and wsk[1] are some random (actually, undefined) int s. They are used as addresses, but they were invalid addresses. Seg fault.

&wsk[0] and &wsk[1] , or more concisely, wsk and wsk+1 , are a valid addresses, where scanf can write an int .

Answer 2

Because you are not passing in pointers to the scanf function in the first case. You are passing in the value of wsk[0] and wsk[1] and treating those values as pointers.

Answer 3

a) Both codes will terminate at least when they are finished.
b) You´re not deleting all reservations before the program ends.
c) You program is a mix of C and C++.
...
d) scanf take addresses to variables. &