Does specifying size of array in parameter have any effect? For example.
void SetIDs(int IDs[22] );
versus
void SetIDs (int IDs[] );
Does it do nothing?
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5It misleads, that's an effect. If you want a specific size of array, pass it by reference. – Cheers and hth. - Alf Sep 15 '15 at 00:23
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1This has been answered, but is a duplicate of http://stackoverflow.com/questions/16144535/difference-between-passing-array-fixed-sized-array-and-base-address-of-array-as – Weak to Enuma Elish Sep 15 '15 at 00:48
1 Answers
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No, if you pass by value, an array type specifier in a parameter is always adjusted to a pointer.
void SetIDs (int IDs[22] );
void SetIDs (int IDs[] );
void SetIDs (int *IDs );
all produce the same code. Even sizeof IDs inside SetIDs will only return the size of a pointer as if IDs was declared as int *IDs.
The array size becomes relevant when you pass by reference or have a multidimensional array:
void SetIDs (int (&IDs)[22] );
This SetIDs will only accept (references to) arrays of int with size 22.
void SetIDs (int IDs[22][42] );
// equivalent to
void SetIDs (int (*IDs)[42]);
This SetIDs will accept pointers to (or an array of) an array of int with size 42.
cg909
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What about program safety? Would the compiler prevent you from passing an int[16] to a function using int[22]? I like having the compiler find errors for me, as it's easier and faster than for me to try and find these issues (passing wrong sized arrays) during runtime. – Thomas Matthews Sep 15 '15 at 01:25
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No, not when passing by value. The array parameter (respectively the first dimension of a multidimensional array parameter) will behave exactly as if it was declared as a pointer, without any warnings or compiler errors. Some compilers like clang will warn e.g. if you use `sizeof` inside the function on an array parameter, but there are no warnings when calling the function with `int[16]` etc. – cg909 Sep 15 '15 at 02:22