Memory address of variables in c

Question

// intialize a char variable, print its address and the next address
char charvar = '\0';
printf("address of charvar = %p\n", (void *)(&charvar));
printf("address of charvar - 1 = %p\n", (void *)(&charvar - 1));
printf("address of charvar + 1 = %p\n", (void *)(&charvar + 1));

// intialize an int variable, print its address and the next address
int intvar = 1;
printf("address of intvar = %p\n", (void *)(&intvar));
printf("address of intvar - 1 = %p\n", (void *)(&intvar - 1));
printf("address of intvar + 1 = %p\n", (void *)(&intvar + 1));

This is a code i found online and here is the concerned output

address of charvar = 0x7fff9575c05f
address of charvar - 1 = 0x7fff9575c05e
address of charvar + 1 = 0x7fff9575c060
address of intvar = 0x7fff9575c058
address of intvar - 1 = 0x7fff9575c054
address of intvar + 1 = 0x7fff9575c05c

My doubt is why the memory address in a computer is stored in a hexadecimal format? We know the size of one char is 8bits or 1 byte, what does 1 byte mean in memory that is the address of the start bit of charvar is 0x7fff9575c05f shouldn't the address of the char+1 be the 0x7fff9575c05f + 8bits be 0x7fff9575c067, but it seems that one memory location in the computer is organised in terms of 8bits or 1 byte. Am i correct?If so why?

Answer 1

The memory is organized in terms of bytes, and pointers point to a specific byte, not to a single bit. The reason is probably that early computers had 8-bit registers/... and usually whole bytes were processed at once. Since the computer was operating on whole bytes, addressing bytes instead of single bits made more sense. It also saves address space, allowing for more memory to be addressed with the same pointer size.

Also the memory addresses are not really stored in hexadecimal format, they are just formatted that way when printed out. Internally in memory they are binary numbers just like all the other numbers a computer works with.

Answer 2

The smallest part of memory you can easily access is a byte, so there would be no use making an address for every bit.

Answer 3

Memory is addressed in bytes rather than bits, you have to accept this as a matter of fact. Once you get a value of a certain byte in a memory you can work with its bits using logical (bitwise) functions or operators, such as &, |, ^, ~ etc. See http://www.cprogramming.com/tutorial/bitwise_operators.html

Moreover, the address is not stored in hexadecimal format, the hexadecimal format is only the format of the number printed to the output. If you used other formatter %d rather than %p in your printf call you will get decadic format. See http://www.cplusplus.com/reference/cstdio/printf/