C++ simple malloc and sizeof

Question

I am getting a weird result with these 2 simple lines

char* reverse = (char*) malloc(sizeof(char)*19);
cout << sizeof(reverse)/sizeof(char) << endl;

No matter what number i put in the first line (in this example, it is 19). I always get 4 as the output. What is wrong ? Thanks.

***What is wrong ?*** You are calculating the size of a pointer. — drescherjm, Dec 29 '16 at 05:05
`char*` and `char[19]` are different types. `sizeof(char*)` is four bytes in your system — Ari0nhh, Dec 29 '16 at 05:05
Don't. Use `std::string` if you want dynamic character arrays. That has a `size` function. — NathanOliver, Dec 29 '16 at 05:07
I am re-implementing several string methods for learning. The duplicate notice explains the other direction. This explains why sizeof function is an exception http://stackoverflow.com/questions/671790/how-does-sizeofarray-work — cuongptnk, Dec 29 '16 at 05:23

Wasi Ahmad · Accepted Answer · 2016-12-29T05:30:37.583

3

On 32-bit machine sizeof pointer is 32 bits (4 bytes), while on 64 bit machine it's 8 bytes. Regardless of what data type they are pointing to, they have fixed size.

And

sizeof(char) = 1 byte

So, you are getting 4 every time because your system is a 32-bit machine.

edited Dec 29 '16 at 05:30

answered Dec 29 '16 at 05:08

Wasi Ahmad

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Your answer would be even better if you explained how to do what the OP was trying to do. – Carey Gregory Dec 29 '16 at 05:09
I think there is no way to recalculate the size – cuongptnk Dec 29 '16 at 05:21
http://stackoverflow.com/questions/10639666/how-to-determine-the-size-of-an-allocated-c-buffer – cuongptnk Dec 29 '16 at 05:21
I agree with both of you. I answered why OP is getting 4 every time and i guess there is no straight forward approach to get the actual size from the pointer (not 100% sure though). If anyone knows that, you can tell me how to do that or you can edit my answer. Thanks :) – Wasi Ahmad Dec 29 '16 at 05:25

C++ simple malloc and sizeof

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