How to split a string into int[3] - c

Question

I have a string, like "101 1 13" and I need to split it to a int aux[3] --> resulting in aux[0] = 101, aux[1] = 1 and aux[2] = 13 (in this case). How can
I do that?
In the example of the code below I get op as a String and want to get the value of the INTs in there. Each int is divided in the string by a white space(" ").
Another detail: I need the code to compile with flag -std=c99, so the answer that was accepted would not work.
#include <stdio.h>
#include <stdlib.h>
//example of str = "101 1 14" (char *)
// example of output = {101, 1, 14}(int *)
int* stoi(char *str) {
// function to split str into 3 ints
}
int main() {
char op[10];
int num[3];
scanf("%s\n", op);
num = stoi(op);
printf("%d %d %d", num[0], num[1], num[2]);
return 0;
}

Answer 1

First you need to tokenize your input (break apart the input into distinct elements). Then you need to parse/integerize the individual tokens by converting them from strings to the desired format.
Sample Code
#include <stdio.h>
#include <string.h>
#define BUF_LEN (64)
int main(void)
{
char buf[BUF_LEN] = { 0 };
char* rest = buf;
char* token;
int i = 0;
int iArr[100] = { 0 };
if ( fgets(buf, BUF_LEN, stdin) != NULL )
{
strtok(buf, "\n"); // Remove newline from input buffer in case we want to call fgets() again.
while ( (token = strtok_r(rest, " ", &rest)) != NULL )
{
iArr[i] = strtol(token, NULL, 10);
printf("Token %d:[%d].\n", i, iArr[i]);
i++;
}
}
return 0;
}
Sample Run
1231 12312 312 1232 1312
Token 0:[1231].
Token 1:[12312].
Token 2:[312].
Token 3:[1232].
Token 4:[1312].

Answer 2

Try to replace your code by following code.
The new code works only if input contains only single space between integers.
Your code:
while(op[cont] != '\0') {
for(i = 0; op[cont] != ' '; i++, cont++) {
num[i] += op[cont];
}
printf("num[i] = %d\n", num[i]);
}
New code:
while(op[cont] != '\0')
{
if(op[cont] != ' ')
num[i] = num[i]*10 + (op[cont]- '0');
else
i++;
cont++;
}

Answer 3

See this example of how to do that:
char string [10] = "101 1 666"
int v [3], n=0, j=0;
int tam = strlen(string);
int current_Len = 0;
for(i=0; i<tam; i++){
//32 = ascii for White space
if(string[i] != 32){
n = n*10 + string[i] - '0';
current_len++;
} else if (current_len > 0){
v[j++] = n;
current_len = 0;
n=0;
}
}
if (current_len > 0){
v[j++] = n;
}

Answer 4

This answer is assuming you know how much integers your string contain at the time of writing your code. It also uses specific clang/gcc extension (typeof) and may not be portable. But it may be helpful to someone (I mainly wrote it because I had nothing good to do).
#include <stdio.h>
#include <string.h>
struct {int _[3];} strToInt3(const char (*pStr)[])
{
int result[3] = {0}, *pr = result;
for(register const char *p = *pStr; *p; ++p)
{
if(*p == ' ') ++pr;
else
*pr *= 10,
*pr += *p - '0';
}
return *(__typeof__(strToInt3(0)) *)result;
}
int main()
{
char op[10];
int num[3];
scanf("%10[^\n]", op),
//memcpy(num, strToInt3(op)._, sizeof(num));
//or
*(__typeof__(strToInt3(0)) *)num = strToInt3(op);
printf("%d %d %d", num[0], num[1], num[2]);
}
I've commented the copying of returned array using memcpy and added a structure assignment. Although both must be valid (not standard I guess but working in most cases) I prefer the second option (and maybe some compiler optimizers will).
Also I assume ASCII character set for chars.

Answer 5

I found an easier approach to the problem. I insert a scanf, that don't catch the space blanket and convert it using atoi. As it is just 3 ints it doesn't become so bad to use this simple, repetitive way of catching the values. And it work with the -std=c99 flag, that I needed to use.
scanf("%s[^ ]\n", op);
num[0] = atoi(op);
scanf("%s[^ ]\n", op);
num[1] = atoi(op);
scanf("%s[^ ]\n", op);
num[2] = atoi(op);
printf("%d\n", num[0]);
printf("%d\n", num[1]);
printf("%d\n", num[2]);