vector<vector<double> >a(3,vector<double>(4));
double *p = a[0];
Why this is wrong, a[0] is not the address of the first dimension of a?
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Bartek Banachewicz
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tenos
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yes, c++11 support it – tenos Jan 14 '14 at 13:26
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5You forgot to ask a question there Skippy – Lightness Races in Orbit Jan 14 '14 at 13:29
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Don't put the answers in the question text. – Bartek Banachewicz Jan 14 '14 at 13:35
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1You've edited the question to remove the question and replace it with (some kind of) answer. If you have an answer, then you could post it as an answer; but in any case you should leave the question intact so the rest of us can understand it. – Mike Seymour Jan 14 '14 at 13:36
1 Answers
4
Look here
vector<vector<double> >a(3,vector<double>(4));
You defined a as a vector having 3 elements of type vector<double>. So a[0] has type vector<double>. vector is a user defined type. It is not a pointer.
Vlad from Moscow
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1You have stated loosely what is wrong. You have not helped explain how the OP can do what they want to do. – thecoshman Jan 14 '14 at 13:26
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@thecoshman: Kinda difficult when the OP didn't tell us what he wants to do. I'd say Vlad hit it straight out of Red Square this time. – Lightness Races in Orbit Jan 14 '14 at 13:31