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c++arraysmemory-address

why is the method of assigning one to another is different from assigning two individual variables in C++?

so in the picture, it says that the problem is with the starting address of the array as we cant change it. but why does this apply only for arrays. int x = 1; we could easily say int y = x; and it would work. doesnt this change the memory address of the variable too?

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Solution

  • [tl;dr] The first and sole reason why newValues = oldValues; is illegal ("will not work") is that the C++ standard prohibits it. Array assignment is not defined, supported or allowed in C++, and therefore any such statement is invalid code. Any other attempts to "explain" it, using memory addresses or other speculations, only obfuscates the simple truth that it is the definition of the language that decides what is legal and what is not in that language.

    The following are quoted from the posted "textbook" excerpt, which I find to be both wrong and misguided for what looks to be an introduction to C++ basics.

    the name of an array without the brackets and subscript stands for the array's starting memory address

    Wrong. The name stands for the variable that it denotes, which has array type. While it is true that an array can decay to a pointer ("starting memory address) in certain contexts, it is certainly not true that an array name is the same with its starting address. For example, both sizeof oldValues and typeid(oldValues) are valid expressions, which would mean something very different if replacing oldValues with its memory address.

    the statement will not work because you cannot change the starting memory address of an array

    The statement "will not work" is correct, but the given reason is still wrong. No assignment changes the address of its left-hand side, it only changes its value. Array assignment does not work because the language does not define it, and for no other reason. Consider for example the following.

    int a[4], b[4];
a = b;  // error, array assignment not allowed

struct { int n[4]; } c, d;
c = d;  // ok, using default copy assignment