How can a C++ array be value-initialized?

Question

1) In C++, is providing the initializer list {} the same as {0}? Will the statements:

int x[10]={};

int x[10]={0};

both produce the same array with all elements initialized to 0?

2) On systems/compilers where NULL is not 0, do default-value initializations of arrays of pointers set the elements to NULL or to 0? Which of the following statements should/can be used?

int *x[10]={NULL};

int *x[10]={0};

int *x[10]={};

How about new value-initializers using empty parentheses -- do they use NULL or 0 as the initializer?

int **x=new int*[10]();

`NULL` is 0 on all compilers that conform to the C++ standard. — dreamlax, May 03 '12 at 03:25
Redefining `NULL` as 1 gave me all 0s for an empty initializer list. — chris, May 03 '12 at 03:26
@chris: Only the preprocessor sees `NULL`, so changing it won't affect how the compiler value-initialises pointers. — dreamlax, May 03 '12 at 03:31
@PreetKukreti: Because that does not always produce useful results (as implementation-defined and unspecified behavior may very across compilers and even with different flags). Best to get a quote from the standard that explains the required behavior. — Martin York, May 03 '12 at 04:00

score 5 · Answer 1 · answered May 03 '12 at 03:28

Yes. Both of those initialisers are equivalent in functionality. The difference is that the second one explicitly initialises the first element to 0, and implicitly value-initialises all remaining elements in the array (which in the case of int, means setting them to 0).
NULL is 0 on all compilers that conform to the C++ standard. In this instance, all three are the same in terms of functionality. NULL is a macro that expands to 0, so the first two are identical in the eyes of the compiler. I'm not sure what the deal will be with C++11, but the value-initialisation of a pointer means setting that pointer to NULL.

How can a C++ array be value-initialized?

1 Answers1