Separating a string into smaller strings - c

Question

I have the following string abcd1234 and I want to find a way to break this string into two different strings, abcd and 1234. I have tried the following code:
char buf[100],*str1,*str2;
int x;
fgets(buf,sizeof(buf),stdin);
str1=strtok(buf,"0123456789 \t\n");
str2=strtok(NULL," \n\t\0");
puts(str1);
puts(str2);
x=atoi(str2);
printf("x=%d", x);
but output is abcd 234. And if I try it with one letter and one number, e.g a2 I take only e on output and x is 0.

Answer 1

As per the man page of strtok()
Each call to strtok() returns a pointer to a null-terminated string containing the next token. This string does not include the delimiting byte. [...]
So, while using "0123456789 \t\n" as the delimiter for the first time, 1 will be treated as the actual delimiter and will not be considered in the subsequent parsing.
You may want to use strcspn() and/or strpbrk() to find out the index for the required sub-strings and parse accordingly.

Answer 2

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
size_t extract(const char **sp, char *out, int (*test)(int ch));
int main(void){
char buf[100], str1[100], str2[100];
int x;
const char *p = buf;
//size_t len;
fgets(buf, sizeof(buf), stdin);
while(*p){
if(isalpha((unsigned char)*p)){
extract(&p, str1, isalpha);
puts(str1);
} else if(isdigit((unsigned char)*p)){
extract(&p, str2, isdigit);
x = atoi(str2);
printf("%s, x=%d\n", str2, x);
} else {
++p;//skip one char
}
}
return 0;
}
size_t extract(const char **sp, char *out, int (*test)(int ch)){
const char *p = *sp;
while(*p && test((unsigned char)*p)){
*out++ = *p++;
}
*out = '\0';
size_t len = p - *sp;
*sp = p;
return len;
}

Answer 3

Try below code.Hope this will help you.
#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main()
{
char string[]="abcd1234";
char digitStr[10];
char charStr[10];
int i,j = 0,k = 0;
for(i=0;string[i];i++)
{
if(isdigit(string[i]))
{
charStr[j++]=string[i];
}
else
{
digitStr[k++]=string[i];
}
}
charStr[j] = '\0';
digitStr[k] = '\0';
printf("%s %s\n",digitStr,charStr);
}

Answer 4

I realize I'm very late on this one, but this is for if anyone has a similar case
Assuming all input strings are like your example, this method will work.
char buf[100];
fgets(buf, sizeof(buf), stdin);
if (buf[strlen(buf) - 1] == '\n')
buf[strlen(buf) - 1] = '\0';
int x = atoi(strpbrk(buf, "0123456789"));
char letters[number - buf + 1];
memcpy(letters, sizeof(letters) - 1, buf);
letters[sizeof(letters) - 1] = '\0';
//letters is the word
//x is the number as an int, not a string
• Note the if statement after the fgets. This checks that the newline character was read by fgets, and turns it into a NUL character. (essentially truncating the string).
• As for strpbrk(), that's just a function that returns a pointer to the first occurence of any character in the second string inside the first string. I use it here to find the start of the digit sequence.
• I would also drop the atoi() for strtol() for safety.
• The letters[] array size is the return of strpbrk() (the address of the first number), minus the start of the array (giving the length of the letter string in bytes), plus one for the NUL character I add later.