void foo (int k)
{
int C[size(k)][size(k)];
C[1][2] = 4;
std::cout << C[1][2];
}
How this code is compiled correctly?
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πάντα ῥεῖ
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David Tsaturyan
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You cannot do this with standard c++ as a raw array. Use a `std::vector
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But it is compiled on g++ correctly. – David Tsaturyan Mar 27 '16 at 11:49
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1GCC has an extensiion to support VLA's (variable length arrays). – πάντα ῥεῖ Mar 27 '16 at 11:50
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size() just return k*2. – David Tsaturyan Mar 27 '16 at 11:53
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@DavidTsaturyan I'll ask it differently. Where do you get size() from? There is no such function in standard C++. – PaulMcKenzie Mar 27 '16 at 11:56
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It isn't standard function. – David Tsaturyan Mar 27 '16 at 11:58
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@DavidTsaturyan Is your question why this actually compiles with g++? – πάντα ῥεῖ Mar 27 '16 at 12:05
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Yes, my question about that – David Tsaturyan Mar 27 '16 at 12:06
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1Then you already have the answer: this is a non-standard gcc extension to the C++ language. – Sam Varshavchik Mar 27 '16 at 12:15
1 Answers
1
You code compiles with g++ because it supports variable length arrays as an extension; see 6.19 Arrays of Variable Length
It's not a feature of standard C++, so it won't necessarily work with other compilers.
If you want g++ to complain about this sort of thing, give it -pedantic.
RichieHindle
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