Size of a string

Hi everybody,

I have a question on calculating the actual size of a string.

Example:

char *string = "String size";
int size;

The actual "String Size" is 11 charactres long.
How can I get the size by using sizeof() routine.

I am trying to use, but that does not work.

size = sizeof(string);

Thanks

Use strlen().

I can't use any Standard Library C functions other than sizeof();

Ah... in that case, you may find sizeof rather useless, methinks. You are dealing with a null terminated string here, so run a loop to count characters until you hit the first null character.

I am going by my C++ background here, but I believe sizeof will just give you the size of the pointer to char in terms of number of bytes.

sizeof("String size")
perhaps?

I am trying to create a MACRO that will do this for a string of any size.
If I have to go with a loop, I may as well just write a routine that does this.
Thanks for your help.

If you declare string like so

Code:

---

char string[] = "String size";

---

then sizeof(string) works.

Well, since the size of a character is 1, you can indeed use sizeof to get the length of a FIXED string (doesn't work if the string is user-input into an array):

Code:

---

char string[] "some value here...";
int sizeOfString = sizeof(string)/sizeof(*string) - 1;

---

The -1 is because of the '\0' which usually isn't part of the string length. Note that this DOESN'T work if the user enters something:

Code:

---

int sizeOfString;
char someString[100];
fgets(someString, 100, stdin);
sizeOfString = sizeof(string)/sizeof(*string) - 1;
printf("Size of the string: %d\n", sizeOfString);

---

This code will print 99, regardless of what the user enters. There is no "other" way except for looping until you find '\0'.

char string[] = "String size";

size = sizeof(string);

This works!

Thanks again!

Quote:

---

Originally Posted by dwks

If you declare string like so

Code:

---

char string[] = "String size";

---

then sizeof(string) works.

---

To elaborate on this, declaring it as a "char *" causes string to be a pointer to a string literal. Taking sizeof will give the size of that pointer (some small value), not the length of the string. Declaring it as "char []" causes it to be an array of chars. Taking sizeof() of an array always gives the size of the array. Since sizeof(char) is equal to 1, this is the same as taking the length of the string. Beware the null terminator; it is included in this count, unlike the return value of strlen().

Quote:

---

Originally Posted by brewbuck

To elaborate on this, declaring it as a "char *" causes string to be a pointer to a string literal. Taking sizeof will give the size of that pointer (some small value), not the length of the string. Declaring it as "char []" causes it to be an array of chars. Taking sizeof() of an array always gives the size of the array. Since sizeof(char) is equal to 1, this is the same as taking the length of the string. Beware the null terminator; it is included in this count, unlike the return value of strlen().

---

Unless it's being passed to a function. This

Code:

---

void function(int array[]) {}

---

is the same as

Code:

---

void function(int *array) {}

---

You can't get a []-style array in a function.

Quote:

---

Originally Posted by dwks

Unless it's being passed to a function. This

Code:

---

void function(int array[]) {}

---

is the same as

Code:

---

void function(int *array) {}

---

You can't get a []-style array in a function.

---

You can if you tell it the size:

Code:

---

void function(int array[6][5]);

---

It works fine.

EDIT: Oh, I see what you mean. Yes, sizeof() won't return the real size in that case. By "[] style array" I assumed you mean with respect to indexing.

This raises the question (that I've heard asked before, but never succintly answered), why isn't there a library function that can return the size of a block of allocated memory? I mean, the OS must keep a table of allocated memory sizes somewhere, and such sizes, either directly or indirectly, are used in conjuction with a call to free() right? So why isn't there a function that can retreive the size of memory allocated given a pointer reference?

size_t _msize( void *memblock );

What exactly is that supposed to be? I'd be interested to see what specific compiler it's for, since it turns up zero hits on Ye Old Popular Search Engine.


Quzah.

Lol. It works in MSVC 6.0. Yea, I know. It's outdated. Check out some code that runs with flying colors in MSVC 6.0. I apologize in advance for casting malloc.

Code:

---

#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>

int main()
{
        int *lpArray = (int*)malloc(sizeof(lpArray) * 101);

        size_t sizeArray = _msize(lpArray) / sizeof(lpArray);

        printf("lpArray has a size of &#37;d.\n", sizeArray);

        return 0;
}

---

[edit]Ye Old Popular Search Engine[/edit]

Ah, my font makes it look like it's all run together: "size_t_msize", which is why I didn't find anything on it.


Quzah.

PellesC, LCC & LCC~Win32 have it in stdlib.h

Code:

---

/* private extensions to standard C */
//...
size_t __cdecl _msize(void *);

---

I found it inside MALLOC.h for the Win32 version of LCC.

Borland supports it, too, under malloc.h.

Can anyone find a popular compiler that does not support it? I'd be interested in knowing how portable this is.

Oh, I don't know...


Quzah.

> I'd be interested in knowing how portable this is.
The fact that it begins with an _ should tell you all you need to know about how useful that function is.

Why is it so hard to store the parameter you passed to malloc in the first place in another variable?

Try using
strlen(String)

Quote:

---

Originally Posted by RakeshKhanna

Try using
strlen(String)

---

I suppose you didn't read the earlier posts?