That is correct except for one extra 1 which seems to be a typo.
This will be the representation of int a = 7, assuming sizeof int is 4 bytes and a little endian machine (start address of memory is from the left).
Why is there a confusion?
No. Its little endian. The least significant byte occupies the first byte of the address. In big endian, the most significant byte occupies the first byte of the address. int a = 7 in big endian will be
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