Okay, I'm working on learning C and working with some memory locations.

Assume the sizeof(double) is 8

and we have a double dbl_array[5] with a memory location of 0x7fffffffe360.

What would the memory location be for &(dbl_array[4])?

When running locally I can see that the location goes from b34b0 to b34d0, but I'm stuck on how to apply this to the assumed location.

Any tips would be amazing!

CodePudding user response：

Calculate the address in bytes as base 4*sizeof(double). So 0x7fffffffe360 4*8.

CodePudding user response：

What would the memory location be for &(dbl_array[4])?

Without knowing the type of computer and/or operating system running this program, this question cannot be answered; it is only possible to say which address this element would have on 99% of all computers:

If X is an array of the data type Y and the address of the element X[n] is Z, the address of the element X[n 1] is Z sizeof(Y).

For this reason, the address of the element X[n M] is Z M*sizeof(Y).

The address of an array is the address of element X[0].

Now simply take a calculator that can calculate hexadecimal numbers and perform the following calculation: 0x7fffffffe360 4 * 8

However, there are counterexamples where the address calculation is done differently: The "huge" memory layout on x86-16 for example...