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# How to convert integer to string array without specify array size?

Hi,

How can I convert integer, for example 12113, to char array but without
specify an array size in C programming?

Thanks!

Dec 7 '05 #1
7 13850
he******@gmail.com wrote:
How can I convert integer, for example 12113, to char array but without
specify an array size in C programming?

You can't. Where would you put it? In random, sizeless memory? There is
no such thing. Whenever you have an array, you will have had to specify
its size somehow.

Richard
Dec 7 '05 #2
Hi Richard,

But how can I read an integer for example, 12113, one by one for
comparison?

Thanks?

Dec 7 '05 #3
he******@gmail.com wrote:
How can I convert integer, for example 12113, to char array but without
specify an array size in C programming?

You can use dynamically allocated memory. To get the size of the string
needed, you can find the log (base 10) of your number, truncate it, and

For example for 12113...

size = floor(log10(12113)) + 1 = 4 + 1 = 5

You can now allocate a string with size characters using malloc and use
it to store your converted string.

Dec 7 '05 #4
why couldn't you just use asprintf ?

Saif wrote:
he******@gmail.com wrote:
How can I convert integer, for example 12113, to char array but without
specify an array size in C programming?

You can use dynamically allocated memory. To get the size of the string
needed, you can find the log (base 10) of your number, truncate it, and

For example for 12113...

size = floor(log10(12113)) + 1 = 4 + 1 = 5

You can now allocate a string with size characters using malloc and use
it to store your converted string.

Dec 7 '05 #5
Ja**********@gmail.com writes:
why couldn't you just use asprintf ?

There is no such thing as asprintf in C.

DES
--
Dag-Erling Smørgrav - de*@des.no
Dec 7 '05 #6
"Saif" <sa****@gmail.com> writes:
he******@gmail.com wrote:
How can I convert integer, for example 12113, to char array but without
specify an array size in C programming?

You can use dynamically allocated memory. To get the size of the string
needed, you can find the log (base 10) of your number, truncate it, and

For example for 12113...

size = floor(log10(12113)) + 1 = 4 + 1 = 5

If I'm going to convert it to an char array I'd use snprintf to
calculate the size:

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

int main(void)
{
char* buf = NULL;
int x = 12113;
size_t sz = snprintf(buf, 0, "%d", x) + 1;
buf = malloc(sz);
snprintf(buf, sz, "%d", x);
/* do something with buf */
free(buf);

return 0;
}
Dec 7 '05 #7
Niklas Norrthon wrote:
If I'm going to convert it to an char array I'd use snprintf to
calculate the size:

snprintf is a new function added in C99 and may not be supported by all
compilers. I know it's not supported in mine.

Dec 8 '05 #8

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