Pointers & converting a char array to an int

Question

I am doing some exercises to figure out how to access values in an array after they are changed with pointers. Can someone point out why the first output does not show the desired output? I am trying to get both cout to print 1234, one by using the new pointer and one by using the position in the array

int main()
{
    char myArray[50]={0};
    short* sizeOfAlloc=(short*)(myArray+5);  
         *sizeOfAlloc=1234;

    cout << (short*)(myArray+5) <<endl; 
    cout << *sizeOfAlloc <<endl;
    system("pause");

}

myArray is a char** but you cast it to a short*. 2 problems: 1. (the main one), you have changed the level of indirection. 2. You have cast from char to short, which you should only do if you really know what you're doing (more likely you want to cast to int16_t) — Dave
– Dave, Commented Apr 4, 2013 at 22:55
Thank you Daniel and Dave. Regarding your first comment dave, do you know of a better way to convert some chars to a double or something else? — user2012481
– user2012481, Commented Apr 4, 2013 at 22:58

user995502 · Accepted Answer · 2013-04-04 22:54:24Z

2

cout << (short*)(myArray+5) <<endl;

Prints the pointer. Not the value pointed by it.

cout << *((short*)(myArray+5)) <<endl;
        ^^                   ^^

Will print the value pointed to by (short*)(myArray+5)

answered Apr 4, 2013 at 22:54

user995502

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3 Comments

Vite Falcon Over a year ago

@stardust_: If this helped resolve your issue, mark this as the answer.

user995502 Over a year ago

@ViteFalcon How? :) he asked the question. Not me :)

Vite Falcon Over a year ago

@stardust_: I meant user2012481. Sorry :)

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