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Is array name a pointer in C?
char arr[1024];
arr++ //move arr to arr+1, error!
I have heard the the name of a char array is a char pointer, is it?
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lovespring
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5 Answers
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The name of an array decays to an pointer to its first element sometimes.
An expression with array type will convert to a pointer anytime an array type is not legal, but a pointer type is.
You cannot do:
arr++;
Because array is an non modifiable l-value.
Alok Save
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sometimes, we can use array name as a pointer (e.g. if a function need a pointer), is this a decay? when we can decay? is there a rule? – lovespring Nov 12 '11 at 08:20
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@lovespring: I stated the rule in my answer. *An expression with array type (which could be an array name) will convert to a pointer anytime an array type is not legal, but a pointer type is.* Example is: Declaration of an array as a function parameter is converted into a declaration of a pointer. – Alok Save Nov 12 '11 at 08:22
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An array is a block of memory to which you gave a name.
What does it mean to "increment" it by one? That doesn't make any sense.
A pointer is a memory address. "Incrementing" it by one means to make it point to the element after it.
Hope that makes sense.
user541686
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Arrays and pointers are not always exchangeable. A more interesting example to illustrate the difference between arrays and pointers is a 2D array:
Consider int **a and int b[3][3].
In the first approach to a 2D array we have a 1D array of pointers to 1D arrays in memory (of course you have to allocate the memory dynamically to use this).
In the second approach of actually using a 2D C array, we have the elements laid out sequentially in memory, and there's no separate location where an array of pointers are stored.
If you try to dereference b, you get a pointer to its first element (i.e. b gets converted to an int (*)[3] type).
Szabolcs
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nothing, apart from looks..
Well and as you defined the array there you declared already 1024 bytes for the array. Also you obviously can't change the array "base".
paul23
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An array name is for most (but not all) purposes identical to a constant pointer (not to be confused with a pointer to a constant). Because it's a constant, you cannot modify it with the increment operator ++. There's a good answer explaining this in more detail in an older similar question.
