FAQ atoi does it work like this (C)

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).

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.

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'.

Quote:

---

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...

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*

> return;

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

Quzah.