# atoi does it work like this

This is a discussion on atoi does it work like this within the C Programming forums, part of the General Programming Boards category; Hi guys, Does atoi work like this: it scans an array element by element 0=49 1=50 2=51 converts these ascii ...

1. ## atoi does it work like this

Hi guys,
Does atoi work like this: it scans an array element by element 0=49 1=50 2=51 converts these ascii values 49 50 and 51 into there digit equivalents and add each one to an identifier/varriable.

Thanks learner(wanting to be a master).

2. No. It doesn't.

x = atoi( "32" );
printf("%d", x ); /* this prints 32 */

atoi( ) takes a string of numbers and translatest it directly into the integer value of the string, as shown above.

Quzah.

3. Okay, let's say this is the program...
Code:
```#include <stdlib.h>
main ()
{
char a[4];
int i;

a = {49, 50, 51, 0};
// 49 = '1', 50 = '2', 51 = '3', 0 = '\0'
// So a is the string "123"
i = atoi (a)
// Now i is the integer value 123
return;
}```
The 0 is very important, as atoi is meant for reading a 0 terminated character string. That is to say, if instead I used a[3] = {49, 50, 51}, then atoi(a) would have just kept reading for integer values in the memory past a[3] because it hadn't yet encounted the string terminator... '\0'.

4. Originally posted by QuestionC
Okay, let's say this is the program...
Code:
```#include <stdlib.h>
main ()
{
char a[4];
int i;

a = {49, 50, 51, 0};
// 49 = '1', 50 = '2', 51 = '3', 0 = '\0'
// So a is the string "123"
i = atoi (a)
// Now i is the integer value 123
return;
}```
The 0 is very important, as atoi is meant for reading a 0 terminated character string. That is to say, if instead I used a[3] = {49, 50, 51}, then atoi(a) would have just kept reading for integer values in the memory past a[3] because it hadn't yet encounted the string terminator... '\0'.
a = {49, .....}???
you try to compile this?
probably not!
any way this is not the meaning of atoi(ascii to integer),

void main(void)
{
char *a;

a = "49";
printf("%d",atoi(a));
}

output:

49 press any key to continue...

5. Sorry, my post was intended to just get the spirit of atoi across as code, not as a program. Here's the code, fixed up to compile, and producting some output.
Code:
```#include <stdlib.h>
main ()
{
char a[4] = {49, 50, 51, 0};
int i;

printf ("a represented as a string is... %s\n", a);
printf ("a represented as an int is... %d\n", a);

// 49 = '1', 50 = '2', 51 = '3', 0 = '\0'
// So a is the string "123"

i = atoi (a);
printf ("atoi(a) as an int is... %d\n", i);
printf ("atoi(a) as a string is... %s\n", i);

// Now i is the integer value 123
return;
}```
Looking at your code... well, here's the thing...
Code:
`char *a = "49";`
is equivalent (kinda) to
Code:
`a[3] = {52, 57, 0};`
and that's pretty much the difference between my code and yours. 52 is the ASCII value for '4', 57 is the ASCII value for '9', and 0 is the string terminator. Similarly
Code:
`char a[4] = {49, 50, 51, 0};`
is the same as
Code:
`char * a = "123"`

*Note to self... use atoi and pointer arithmatic to perform divisions in next ofuscated code contest*

6. > return;

Just to nit pick...you forgot to return a value!

Quzah.