length of char pointer c++ [duplicate]

Question

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

When declaring a char pointer using malloc for example. is there a way to determine the allocated length.

Example, if I define a char array, I can easily find the number of elements

char a[10];
a[0]='c';
cout << sizeof(a) / sizeof(a[0]);

which would yield 10. on the other hand if I tried with a char pointer

char *a;
a = (char*)malloc(10);
a[0] = 'c';
cout << sizeof(a) / sizeof(a[0]);

then it yields 8, which is obviously the size of the pointer, not the allocated length.

is it possible to find the allocated length of a char pointer?

The reason I'm asking this and not use std::string or other, is because I came upon this question while cracking some interview question. so its out of curiousity

IS this C question, not C++? Maybe change the tag, at least add C tag, C folks have a plenty of solution to this problem. In C++ we use std::string. — PiotrNycz
– PiotrNycz, Commented Oct 5, 2012 at 15:51
you are quite right. I edited the tag. its an interview question, I edited my question btw — Moataz Elmasry
– Moataz Elmasry, Commented Oct 5, 2012 at 16:02

user529758 · Accepted Answer · 2012-10-05 15:56:38Z

6

SomeRandomGuyOnSO:

No it's not.

Me:

Yes, it is, but it's gonna be non-standard and platform dependent.

Apart from that, let me not repeat all the good advice about using standard C++ containers; let me actually answer your question instead. Doing a quick Google search on the topic brings up methods for doing this. Examples:

// OS X
#include <malloc/malloc.h>

char *buf = malloc(1024);
size_t howManyBytes = malloc_size(buf);
printf("Allocated %u bytes\n", (unsigned)howManyBytes);

// Linux
#include <malloc.h>

char *buf = malloc(1024);
size_t howManyBytes = malloc_usable_size(buf);
printf("Allocated %u bytes\n", (unsigned)howManyBytes);

// Windows, using CRT

char *buf = malloc(1024);
size_t howManyBytes = _msize(buf);
printf("Allocated %u bytes\n", (unsigned)howManyBytes);

All three snippets are supposed to print 1024 or greater.

answered Oct 5, 2012 at 15:56

user529758

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6 Comments

tomahh Over a year ago

I didnt document myself enought, sorry about the misleading post.

user529758 Over a year ago

@TomAhh No problem, I'm not angry of you in any means - a bit more experience and you'll see that there is rarely an "impossible" thing in programming :)

Moataz Elmasry Over a year ago

@H2CO3 thanks looks like a very nice solution. you are using 1024 for size, is there a reason for that, or can it be any arbitary size for your functions to work?

user529758 Over a year ago

@MoatazElmasry I'm using it because it's nice.

WhozCraig Over a year ago

+1: Direct answer to the question, and directly notes the answer it is platform dependent. Marked as one of my favs. I wouldn't use it in production code, but for trace-debugging this is outstanding. Thanks.

|

Herms · Accepted Answer · 2012-10-05 15:46:17Z

2

malloc() doesn't really know about arrays or other things, it just allocates a block of memory for you. The size of this space isn't available to you unless you track it yourself separately. This is why most methods that act on arrays take two arguments, the array and the length.

answered Oct 5, 2012 at 15:46

Herms

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