converting an int to char*

Question

This is a very very basic question and I know one way is to do the following:

char buffer[33];
itoa(aq_width, buffer,10);

where aq_width is the int, but then I can't guarantee what size of buffer I would need in order to do this... I can always allocate a very large buffer size, but that wouldn't be very nice... any other pretty and simple way to do this?

Answer 1

std::stringstream ss;
ss << 3;
std::string s = ss.str();
assert(!strcmp("3", s.c_str()));

Answer 2

You can calculate an upper-bound on the required size of the buffer using this macro (for signed types):

#define MAX_SIZE(type) ((CHAR_BIT * sizeof(type) - 1) / 3 + 2)

By the way, itoa() isn't standard C and isn't available everywhere. snprintf() will do the job:

char buffer[MAX_SIZE(aq_width)];
snprintf(buffer, sizeof buffer, "%d", aq_width);

Answer 3

A nice function from glibc is asprintf , used like

char * buffer = NULL;
int buffer_size = asprintf(&buffer,"%d",aq_width);
assert(buffer_size >= 0);
...
free(buffer)

which will allocate the buffer for you.

More portably you can use snprintf , which will return the number of characters you need to allocate.

Answer 4

There are more memory efficient ways of doing it, but they aren't "simple".

Consider the cost of 100 bytes of memory for, say, one second. A gigabyte for lifetime of ten years costs, what, $10? We're talking nanocents here.

Make the buffer too big. Note that a 32-bit int can't be more than 10 digits in decimal, and a 64-bit int, can't be more than 20.