sizeof works unusual at c++ after array creation [duplicate]

Question

Possible Duplicate:
How to find the sizeof(a pointer pointing to an array)

I am creating an array by using following code

float *A;
A = (float *) malloc(100*sizeof(float));
float *B;
B = (float *) malloc(100*sizeof(float));

but after these when I type an print the size of the A and B by the following, I get 2 as a result as I expect to see 100.

sizeof(A)/sizeof(float)

Your expectation, that the size of a pointer should change if you change what it's pointing to, is unreasonable. (Now, if this was an array passed to a function, you'd have a point.) — David Schwartz
– David Schwartz, Commented Jan 27, 2013 at 0:40

Luchian Grigore · Accepted Answer · 2013-01-27 00:27:53Z

11

Your expectation is wrong. A is a float*, so its size will be sizeof(float*), regardless of how you actually allocate it.

If you had a static array - i.e. float A[100], then this would work.

Since this is C++, use std::array or std::vector.

Worst case, use new[]. Definitely don't use malloc.

answered Jan 27, 2013 at 0:27

Luchian Grigore

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12 Comments

Luchian Grigore Over a year ago

@Erogol there would be if you took good advice and stop writing C code and labeling it C++. Use vector or array.

Luchian Grigore Over a year ago

@Erogol so, wait, do you want to learn C or C++? You do know they're different, right?

Etienne de Martel Over a year ago

@Erogol If you tag your question C++, expect C++ answers.

Cat Plus Plus Over a year ago

@Erogol Sorry, you can't have an account here if you're under 13.

Cat Plus Plus Over a year ago

@Erogol Stop throwing tantrums like a spoiled stubborn child. You got your answer already. Annoy and insult more people who try to help you even though they don't have to, see how well that goes~

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nameless · Accepted Answer · 2013-01-27 00:30:01Z

4

This works only for static arrays, defined in the current scope.

All you get in your example is the size of a pointer to float divided by the size of float.

answered Jan 27, 2013 at 0:30

nameless

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