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Editorial Team
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Editorial Team
Asked: May 11, 20262026-05-11T18:30:00+00:00 2026-05-11T18:30:00+00:00

Basically I would like to store the address of a pointer in a buffer.

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Basically I would like to store the address of a pointer in a buffer. Don’t ask me why

char * buff = "myBuff";
char * myData = (char*)malloc(sizeof(char*));
int addressOfArgTwo = (unsigned int)buff;
memcpy(myData, &addressOfArgTwo, sizeof(char*));

cout << "Int Val: " << addressOfArgTwo << endl;
cout << "Address in buffer:" << (unsigned int)*myData << endl;

I can’t see why the above code doesn’t work. It outputs:

Int Val: 4472832
Address in buffer:0

When the Int Val & Address in Buffer should be the same. thanks

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  1. Editorial Team
    Editorial Team
    2026-05-11T18:30:01+00:00Added an answer on May 11, 2026 at 6:30 pm

    You dereference a char *, resulting in a char, and then cast that 1-byte char to an int, not the entire 4 bytes of address (if this is a 32-bit machine, 8 bytes on 64-bit). 4472832 is 444000 in hexadecimal. On a little-endian machine, you grab that last 00.

    *((unsigned int*)myData)
    

    should result in the correct number being displayed.

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