GDB question with C program?

Memory is byte addressable meaning each byte has a distinct address. Then you can interpret say a pair of bytes as a 16-bit integer, four bytes as a 32-bit integer, etc. That format specifier lets you do just that - print content of memory according to your interpretation of it.So x/2x 0x7fff5fbff948 will print two word-size (here 64-bit) integers, first at address 0x7fff5fbff948 and second at address 0x7fff5fbff950.

Memory is byte addressable meaning each byte has a distinct address. Then you can interpret say a pair of bytes as a 16-bit integer, four bytes as a 32-bit integer, etc. That format specifier lets you do just that - print content of memory according to your interpretation of it.So x/2x 0x7fff5fbff948 will print two word-size (here 64-bit) integers, first at address 0x7fff5fbff948, and second at address 0x7fff5fbff950.

First, thank you for your reply. Okay but two 64 bit integers clearly corresponds to two 8 byte integers. Since each byte has its distinct address, if I do x/2x and print two 64bit integers at addresses ____48 and ____50, that means there should be overlap with _____48 and _____50, and I don't see any overlap in the third byte of 48 and the first byte of 50 and thenon.

Can you help explain this? – aengle1429 May 24 at 1:16 Why would there be overlap? 0x50 - 0x48 = 8, i.e.

Eight bytes. – Nikolai N Fetissov May 24 at 1:33 :) Thank you. Stupid mistake... forgot we're in base 16.

– aengle1429 May 24 at 1:49.

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