is the name of an array an rvalue? [duplicate]

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Possible Duplicate:
Is array name a pointer in C?

#include <stdlib.h> int main(int argc, const char *argv[]) { char *b=(char*)malloc(sizeof(char)*50); b=(char*)"hello world"; // works char a[50]; a=(char*)"hello world"; //doesn't work. why? I thought array names are just pointers that point //to the first element of the array (which is char). so isn't a char*? return 0; }

I think the reason it doesn't work is because there's no variable called "a" that actually stores a char* value. so should 'a' be considered an rvalue? I'm not sure if I'm understanding the concept correctly

Arrays are not pointers, yet you seem to be trying to use it as such. — Qaz
– Qaz, Commented Jun 6, 2012 at 16:18
There's a very thorough explanation of these things (and more) here. — R. Martinho Fernandes
– R. Martinho Fernandes, Commented Jun 6, 2012 at 16:19
Be aware that your first example compiles, but it doesn't "work"! It doesn't copy "hello world" into the allocated space, it just changes b to point to the "hello world" string. — TonyK
– TonyK, Commented Jun 6, 2012 at 16:23
Warning: the tags affect the answers. For example, Als answers for C when he says that passing an array name to a function causes decay to pointer (it doesn't in C++ for a function that takes a reference-to-array). But "rvalue" is a term from C++, not from C, which instead says "the value of an expression". — Steve Jessop
– Steve Jessop, Commented Jun 6, 2012 at 16:36

Alok Save · Accepted Answer · 2012-06-07 05:37:57Z

8

Arrays are not pointers, sometimes^{[Note 1:]} the name of an array decays to a pointer when array name is not valid(eg: passing to function).
Arrays are non modifiable l-values, they cannot be assigned and there address can be taken.

^{[Note 1:]}
For example:
Array name doesn't decay to a pointer when used in sizeof()

Array address cannot be changed but content can be changed.

edited Jun 7, 2012 at 5:37

user1004682

answered Jun 6, 2012 at 16:19

Alok Save

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3 Comments

Steve Jessop Over a year ago

@Luchian: "compile-time" in C89, not for C99 VLAs. It's always an operator, not a function.

Alok Save Over a year ago

@SteveJessop: Bang on. For VLA's standard mandates sizeof does run-time evaluation.

legends2k Over a year ago

@AlokSave Arrays are non modifiable l-values, they cannot be assigned -- generally true but there's an exception to this rule too: char const str[] = "this will be copied"; The string literal is a lvalue char array that is copied to str.

AnT stands with Russia · Accepted Answer · 2012-06-06 16:22:43Z

Array is a non-modifiable lvalue. You cannot assign anything to it, yet you can apply the unary & operator to it.

You are right when you say that there's no char * variable involved here. Array name directly refers to an array object - a continuous block of memory whose size is equal to the product of the element count and element size.

Luchian Grigore · Accepted Answer · 2012-06-06 16:19:27Z

I thought array names are just pointers

No they're not. They're arrays. The decay into pointers when you pass them as parameters, but that's about it. You can't re-assign an array, you can only change individual values.

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