convert c++ code to c# in declaring an array

Question

recently i want to convert some codes from opencv to c#. But I have some problem in declaring the dynamic array in c++.

int* memorySteps = 0;
CV_CALL( memorySteps = new int [ 10*sizeof(memorySteps[0]) ]);

I want to convert it to a normal c# int array. The problem is the length.In c++, the declaration above seems that the compilor will create an array with 40 int elements(coz the sizeof the type memorySteps[0] is 4, 4 bytes length.). Is my understanding correct? However, it doesnt make sense because actually in the algorithm I only need an array of 10 int elements. So can anyone give me a hint which is right in the following(in c#), or can anyone show me other examples:

int[] memorySteps =new int [10*sizeof(int)];

or

int[] memorySteps =new int [10];

Welcome to the wonderful world of finding bugs in someone else's code! — Sergey Kalinichenko
– Sergey Kalinichenko, Commented Mar 8, 2012 at 14:41

Mike Seymour · Accepted Answer · 2012-03-08 14:44:55Z

2

Your understanding is correct; the C++ version will create an array of 40 integers, assuming sizeof(int)==4. It may be that the author was getting mixed up with C's malloc, which takes a size in bytes.

So your first version will give the same behaviour, and the second is more likely to give the intended behaviour - but there is a danger that the extra-large buffer was hiding other bugs, which might suddenly be exposed if you reduce the size.

answered Mar 8, 2012 at 14:44

Mike Seymour

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1 Comment

fskking Over a year ago

I wrote a loop to check the length of the dynamic array. for(int n=0;memorySteps[n];++n ){} . the result of n should be (in my understanding) 40, but it returns the result of 44. It really confuses me.

Sergey Kalinichenko · Accepted Answer · 2012-03-08 14:43:42Z

2

If the algorithm needs ten items, then use

int[] memorySteps =new int [10];

I suspect that the C++ code was itself ported from C code, where the programmer called malloc, like this:

memorySteps = malloc(10*sizeof(memorySteps[0]));

Whoever converted that code to C++ has probably forgotten to take away the multiplier. This has resulted in a functionally correct, yet somewhat memory inefficient code (although wasting 30 bytes these days hardly even counts as inefficient).

answered Mar 8, 2012 at 14:43

Sergey Kalinichenko

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Comments

breaker · Accepted Answer · 2012-03-08 14:44:03Z

0

What you need is:

int[] steps = new int [10];

answered Mar 8, 2012 at 14:44

breaker

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