Reading a string and store it in multiple arrays in C

[omardanniel]

Hello I'm new to C and I'm trying to take a string input that takes up to 3 words, and store it in one array, and then store every word from that input in another 3 different arrays, and return the number of inputs. Here it's what I have:(note, I don't want to use 2D or 3D arrays)

#include<stdio.h>
#include<string.h>


void breakUp(char *, char *, char *, char *, int );
int main()
{
 char line[80], comm[10], p1[10], p2[10];
 int len, n;

 printf("Please Enter a command: ");

 fgets(line, 80, stdin);
    /* get rid of trailing newline character */
    len = strlen(line) - 1;
    if (line[len] == '\n')
        line[len] = '\0';
    
    /* Break up the line */
    breakUp(line, comm, p1, p2, &n);
    
    printf ("%d things on this line\n", n);
    printf ("command: %s\n", comm);
    printf ("parameter 1: %s\n", p1);
    printf ("parameter 2: %s\n", p2);
    
    return 0;
}

void breakUp(char *str, char *c, char *p1, char *p2, int n)
{
    c[0] = p1[0] = p2[0] = '\0';
    int j = 0;
    int i = 0;
    
    while(str[j]!= ' ')
      {
	c[i]= str[j];
	i++;
	j++;
      }
    if (str[j] == ' ')
      {
	n++;
	j++;
	i =0;
      }
    while(str[j] != ' ')
      {
	p1[i]= str[j];
	j++;
	i++;
      }
    if(str[j] == ' ')
      {
	n++;
	j++;
	i = 0;
      }
    while(str[j] != ' ')
      {
	p2[i] = str[j];
	j++;
	i++;
      }
    if(str[j] == ' ' || str[j] == '\n')
      {
	n++;
	  }

}

[Thomas1965]

What is the problem ?

[jonnin]

\n isn't always 1 char; it is 2 on windows... use 13 or 10 ascii value maybe.
should be fairly straightforward:


strcpy (worker, all3input);
int x = strlen(worker);
for(j = 0; j< x; j++)
if(worker[j] == ' ')
worker[j] = 0;

strcpy(res1, worker);
worker += strlen(res1);
strcpy(res2, worker);
worker += strlen(res2);
strcpy(res3, worker);

something like that. You will need worker to be a pointer into a string. You will have to double check my pointer math, I could be off by one as I typed that in a hurry. But the algorithm/approach should do it.

if they put in more than 3, it was unclear what you wanted to do.

[omardanniel]

For this problem I'm not allowed to use any of the string functions, the problem right now is that I can't return the number of words entered to my 'n' variable, and when I print the words I can't get the second word to show up.

[JLBorges]

> \n isn't always 1 char; it is 2 on windows...

In every C and C++ implementations, '\n' is required to be a single (unique) character.
Though it may be represented externally, outside the program (for example in text based internet protocols like HTTP or in a file system) as some implementation specific sequence of two or more bytes.

The C programming language provides the escape sequences '\n' (newline) and '\r' (carriage return). ... The C standard only guarantees two things:

1. Each of these escape sequences maps to a unique implementation-defined number that can be stored in a single char value.
...
https://en.wikipedia.org/wiki/Newline#In_programming_languages

#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <assert.h>

// extract the next token in srce into token.
// return the number of characters processed in srce
// invariant: non-NULL pointers
size_t extract_next_token( const char* srce, char* token )
{
    size_t n = 0 ;

    while( *srce && isspace(*srce) ) { ++srce ; ++n ; } // skip leading white space
    for( ; *srce && !isspace(*srce) ; ++n ) *token++ = *srce++ ; // copy non-ws characters
    *token = 0 ; // null terminate

    return n ; // number of characters processed
}

// extract up to a maximum of sz tokens. return number of tokens extracted
// invariant: non-NULL top-level and non-top-level pointers
size_t extract_tokens( const char* srce, char* tokens[], size_t sz )
{
    for( size_t i = 0 ; i < sz ; ++i )
    {
        size_t offset = extract_next_token( srce, tokens[i] ) ;
        if( tokens[i][0] == 0 ) return i ; // return number of tokens extracted
        srce += offset ;
    }

    return sz ;
}

// extract up to three tokens into a, b and c. return number of tokens extracted
// invariant: non-NULL pointers
size_t break_up( const char* srce, char* token_a, char* token_b, char* token_c )
{
    char* tokens[3] = { token_a, token_b, token_c } ;
    return extract_tokens( srce, tokens, 3 ) ;
}

int main()
{
    const char text[] = "   C is quirky, flawed,  \t  and an enormous success.\n"
                        "       - Dennis M. Ritchie.\n         " ;

    enum { MAX_TOK_SZ = 128, NUM_TOKS = 16 };
    char tokens[NUM_TOKS][MAX_TOK_SZ] ;

    char* tok_ptrs[NUM_TOKS] ;
    for( size_t i = 0 ; i < NUM_TOKS ; ++i ) tok_ptrs[i] = tokens[i] ;
    size_t n = extract_tokens( text, tok_ptrs, NUM_TOKS ) ;
    for( size_t i = 0 ; i < n ; ++i ) printf( "%s ", tokens[i] ) ;

    char a[MAX_TOK_SZ], b[MAX_TOK_SZ], c[MAX_TOK_SZ] ;
    n = break_up( text, a, b, c ) ;
    assert( n == 3 ) ;
    printf( "\n\n1. %s\n2. %s\n3. %s\n", a, b, c ) ;
}

http://coliru.stacked-crooked.com/a/6acfcd21e6c0ac15