Please explain me this c language code.

Question

Void f(int* x){
Printf("%d",(*x)++);
*x=2;
}
Int main(){
Int a = 3;
Int* b = &a;
f(&a);
Printf("%d",*b);
Return 0;
}

Answer 1

Hi.

C is entering the code to execute at the main-function. First thing it does is to reserve memory for an integer value called a and assign the value 3 to it.
Next, a pointer to an integer memory is reserved called b and it points to the memory of integer a. Then function f is called which takes an integer pointer as input value called x. In function f the value of contained at the memory of x (asterisk * is the notation to receive the value of the memory it points to) is incremented (with no effect) and printed to command line. After that it is set to 2 and returning to the execution in main. Here the value contained at memory position is printed, which should be 2 due to the same memory as x pointed to before. Return 0 to show that all went fine with your execution.

Hope that helps. Cheers. C.

Post scriptum: Be aware of case sensitivities. All Int should be int and so on. There will be no new line or whitespace between the two printf

Answer 2

Thanks for your explaination
But tell me what is the output

Answer 3

See here

https://code.sololearn.com/cdoPN4MAr5kc/?ref=app

I only change the cases of some keywords