如何解释char * x [5]声明？

Question

My professor has a bit of code uploaded online, but I'm finding it hard to understand this variable:

char * x[5];

Does this represent 5 spaces x has for pointers to characters? So if I were to say

x[0]="Apple";

would this be valid?

Answer 1

Does this represent 5 spaces x has for pointers to characters?

Yes. One way to examine this is to evaluate sizeof(x) , sizeof(x[0]) , and sizeof(*x[0]) . On my x64 machine, those evaluate to 40, 8, and 1 -- pointers are 8 bytes long, so char* x[5] declares five pointers.

would x[0]="Apple"; be valid?

Yes, although it might not do what you expect. Here, x[0] is being set to point to a string literal, and the values of x[1] through x[4] are undefined. Note that because "Apple" is a string literal, it's read-only; you're not guaranteed to be able to modify it.

Answer 2

Yes that what it is, ie an array of 5 char pointers, and indeed

x[0] = "Apple";

is valid.