The C Programming Language, 2nd Edition, by Kernighan and Ritchie
Exercise 1.12 on page 21
Write a program that prints its input one word per line.
Solution by Vidhan Gupta
/* Write a program that prints its input one word per line. */ #include <stdio.h> #define IN 1 #define OUT 0 int main() { int c, state; state = OUT; while ((c = getchar()) != EOF) { if (c == ' ' || c == '\t' || c == '\n') state = OUT; else if(state == OUT){ state = IN; putchar('\n'); } putchar(c); } return 0; }
OUTPUT: What the hell is this What the hell is this
Solution by Richard Heathfield (corrected by FF)
#include <stdio.h> int main(void) { int c; int inspace; inspace = 1; while ((c = getchar()) != EOF) { if (c == ' ' || c == '\t' || c == '\n') { if (inspace == 0) { inspace = 1; putchar('\n'); } /* else, don't print anything */ } else { inspace = 0; putchar(c); } } return 0; }
Solution by Dabeau
#include <stdio.h> #define IN 1 /* inside a word */ #define OUT 0 /* outside a word */ main() { char c ; short state ; state = OUT ; while ((c = getchar()) != EOF) { if (c != ' ' && c != '\t' && c != '\n') { if (state == OUT) state = IN ; putchar(c) ; } else if (state == IN) { state = OUT ; putchar('\n') ; } } }
Solution by Martijn Vruggink
#include <stdio.h> // print input one word per line int main() { int c, prev; while ((c = getchar()) != EOF) { if (c == ' ' || c == '\n' || c == '\t') { if(prev != ' ' && prev != '\n' && prev != '\t') //insert a newline if c is a "non-word" character, prev is a "word" character { putchar('\n'); } } else { putchar(c); } prev = c; } return 0; }
Solution by Chrismath
//ex1-12 #include <stdio.h> #define IN 1 #define OUT 0 // print input one word per line int main(void) { int c, state; // start without a word state = OUT; while ((c = getchar()) != EOF) { // if the char is not blank, tab, newline if (c != ' ' && c != '\t' && c != '\n') { // inside a word state = IN; putchar(c); // otherwise char is blank, tab, newline, word ended } else if (state == IN) { state = OUT; putchar('\n'); } } return 0; }
Solution by Timrobinson
Timrobinson's solution [Warning: This is buggy and fails the "one word per line" part of the spec. Run it on its own source - Do'h! This "solution" should be removed. There is a reason that received solutions track "inside word" or "in space" state. Failure to grok why that is necessary is why this code fails. (unsigned comment by Nick)]
//ex1-12 #include <stdio.h> // print input one word per line. Note: no need to track whether "in" or "out" of word. int main(void) { int c; while ((c = getchar()) != EOF) { // if the char is not blank, tab, newline if (c != ' ' && c != '\t' && c != '\n') putchar(c); // otherwise char is blank, tab, newline, word ended else putchar('\n'); } return 0; }
Solution by Tomi
Another solution:
#include <stdio.h> int main (void) { int c,c_previous; while ((c=getchar()) != EOF) { if (!( (c_previous == ' ' || c_previous == '\t' || c_previous == '\n') && (c == ' ' || c=='\t' || c == '\n'))) { if (c==' ' || c=='\t') { putchar('\n'); } else putchar(c); } c_previous = c; } return 0; }
Solution by akw
#include <stdio.h> main () { int c; while((c = getchar()) != EOF) { if (c != ' ' && c != '\n' && c != '\t') { while (c != ' ' && c != '\n' && c != '\t' && c != EOF) { putchar(c); c = getchar(); } putchar('\n'); } } }
Solution by KreativitetNO
An alternative solution that uses arrays which are discussed in the next section after the exercise:
#include <stdio.h> main() { int c, bp; char buffer[1024]; bp = 0; while ((c = getchar()) != EOF) { /* any word separating character you define goes here */ if (c != ' ' && c != '\t' && c != '\n') { buffer[bp++] = c; else if (bp > 0) { printf("%s\n", buffer); bp = 0; } buffer[bp] = '\0'; } if (bp > 0) printf("%s\n", buffer); }
Solution by Indianlamp
Here is my solution (sorry but this solution is utterlly wrong!!! Have you tested it?Try enter spaces or tabs first [unsigned comment added by Takumi08]):
int main() { int c; while ((c = getchar()) != EOF){ putchar(c); if (c == ' ' || c == '\n' || c == '\t'){ printf("\n"); /* or putchar('\n'); */ } } return 0; }
Solution by Lvictor
How about this:
#include <stdio.h> int main(){ int c,d; d = 0; while((c = getchar()) != EOF){ if(c != ' ' && c != '\n' && c != '\t') d = 0; if(d == 0) putchar(c); if(c == ' '|| c == '\t'){ d = d + 1; if(d == 1){ putchar('\n'); } } } }
Solution by Alban Kurti
#include <stdio.h> #include <stdlib.h> #define OUT 0 #define IN 1 int main() { int c, state; state = OUT; while ( ( c = getchar() ) != EOF){ if ( state == IN && c == ' ' || state == IN && c == '\t' || state == IN && c == '\n' ){ state = OUT; printf("\n");} if ( c != ' ' && c != '\t' && c != '\n' ){ state = OUT;} if ( state == OUT && c != ' ' && c != '\t' && c != '\n' ){ putchar(c); state = IN;} } return 0;}
Solution by Takumi08
#include <stdio.h> /*write a program that prints its input one word per line*/ int main(void) { int c,charPrev; c=charPrev=0; while((c=getchar())!=EOF) { if(charPrev==0) { if(c!=' '&&c!='\t'&&c!='\n') { putchar(c); charPrev=1; } } else { if(c!=' '&&c!='\t'&&c!='\n') putchar(c); else { putchar('\n'); charPrev=0; } } } return 0; }
Solution by sl4y3r 0wn3r
/* * * Exercise 1-12. Write a program that prints its input one word per line. * */ #include <stdio.h> #define IS_UPPER(N) ((N) >= 'A' && (N) <= 'Z') /* 'A'==65 && 'Z'==90 */ #define IS_LOWER(N) ((N) >= 'a' && (N) <= 'z') /* 'a'==97 && 'z'==122 */ #define IS_ALPHA(N) (IS_LOWER(N) || IS_UPPER(N)) /* [A-Za-z] x*/ #define OUT 0 #define IN 1 int main(void) { int c = EOF, state = OUT; while ((c = getchar()) != EOF) { if (IS_ALPHA(c)) { state = IN; putchar(c); } else if (state == IN) { state = OUT; putchar('\n'); } } return 0; }
Solution by Dylan Falconer
#include <stdio.h> #define IN 1 #define OUT 0 int main(void) { int c, state; state = OUT; while ((c = getchar()) != EOF) { if (c == '\n' || c == ' ' || c == '\t') { if (state == IN) { state = OUT; putchar('\n'); } continue; } state = IN; putchar(c); } }
