int minimum(int a, int b, int c){
  int mini =a*b*c;
  int iterator=0;
  int test[3];
  test[0]=a;
  test[1]=b;
  test[2]=c;
  for (iterator=0;iterator<3;iterator++){
    if (test[iterator]<mini){
      mini=test[iterator];
    }
  }
  return mini;
}

I hope that your 3 numbers aren't larger than 1290!

By Another random student of algorithms., 2017-12-14 01:28:18
#include<stdio.h>
#include<stdlib.h>	
#include<string.h>	
	int main(void){
	void run_func(char *map[], char *fnt  , char *km , float *x ,  float *y , int line , int row , int lift    );
				char *long_met[3][8] = {{"mile","0.621371"  ,  "yard","1093.61" , "fut","3280.84" , "duim","39370.1" } ,
							{ "mile" ,"1.60934" , "yard","0.0009144" ,  "fut","0.0003048" ,  "duim","0.0000254"},
 			   				{"kilometer","1"  , "meter","1000"  ,   "stmeter","100000" ,   "mmeter","1000000" } }; 
				char *amount[3][4] = { { "gallon" , "0.264172" ,       "quarta" , "1.05669"     } , 
						{"gallon" , " 3.78541"  ,     "quarta" , "0.946353"   },
						{"litr" , "1"  , "mililitr" , "1000"   }};
				char *mass[3][8] = {   {"eng.tonna","0.984207"  ,  "amer.tonna","1.10231" , "stone","157.473" , "funt","2204.62" } ,
   						{ "eng.tonna" , "1.01605" , "amer.tonna", "0.907185" ,  "stone","0.00635029" ,  "funt","0.000453592"},
						{"tonna","1" ,         "kilogram" , "1000"  , "miligram","100000"   , "microgram","1000000"  }}; 
				char **cp;
				char *buf_data;
				char *fnt_sys;
				char *mtr_sys;
				char *word[100];
	while(1){
	int bg  = 0,convert_ch = 3,y = 0, d = 0, numb = 0;
  	float mn =0  , xm = 0 ;
	printf("%s", "enter data for converter: ");
		fgets( (char *) word, 99 ,stdin);
		buf_data = strtok((char *) word, " ");
		if( ! strcmp(buf_data, "funt.sys"   )     ){
		convert_ch = 0;	
		}
		else	if( ! strcmp(buf_data, "metric.sys"    )     ){
			convert_ch = 1;	
		}
		for( bg = 0 ;  buf_data != NULL;  buf_data = strtok(NULL, " ")  , bg++ ) {
			switch(bg){
				case 1:
					if( !strcmp("long_met" , buf_data  )  ){
			  			 y	=   sizeof(*long_met) / sizeof(long_met[0][0]); 
	          	      	                 d	=   sizeof(long_met) /   sizeof(long_met[0][0]); 
			  		        cp  = &long_met[convert_ch][0]  ;
					}
			   		else if(!strcmp("amount" , buf_data  ) ){
			  			 y	= sizeof(*amount) /  sizeof(amount [0][0]); 
		            	                 d	= sizeof(amount) /  sizeof(amount[0][0]); 
				    	        cp = &amount[convert_ch][0] ;
					}
				 	else if(!strcmp("mass" , buf_data  ) ){
				  		 y	= sizeof(*mass ) /  sizeof(mass[0][0]); 
		         		         d	= sizeof(mass ) /  sizeof(mass[0][0]); 
				 		 cp = &mass[convert_ch][0] ; 
					}
					break;
				case 2:
				fnt_sys = buf_data;
					break;
				case 3:
				mtr_sys = buf_data;
					break;
				case 4:
				numb =   atoi(buf_data);
					break;
			}
		}
		if( !y || !d || !cp || convert_ch == 3 || !numb ){
	
			puts("error");
		}
		else{
			run_func( cp,   fnt_sys ,  mtr_sys , &mn ,  &xm , y  , d , convert_ch );
			if( !mn || !xm  ){
		puts("error");
				} else{
			printf("%f\n" , !convert_ch  ?    (mn  /  xm ) * numb :  (  mn  *  xm   ) * numb         );	
			}
		}
	}
	return 0;
	}
	void run_func(char *map[], char *fnt  , char *km , float *x ,  float *y , int line , int row  , int lift    ){
		int m ;
	if(  ( lift )   ){
			row    -= line;	
		}
	for(  m = 0 ; m <= line   ; m++){
		if(!strcmp(fnt,map[m])){
			*x = atof(map[m  + 1] )  ;
			break;
			}
		}
	for( m =   (row  - line) ; m <=  row - 1  ; m++){
		if( !strcmp( km   ,  map[m] ) ){
			*y = atof(map[m  + 1 ] );
			break;
				}
			}
		}
By Gne4do, 2019-02-26 14:38:14
case ClientMessage:    
    if (*XGetAtomName(GLWin.dpy, event.xclient.message_type)
        == *"WM_PROTOCOLS")
        {   printf("Exiting sanely...\n");
            done = True;
        }
    break;

someone just want to watch the world burn

By Anonymous, 2018-01-07 13:01:05
strcpy(szBuffer, sBuffer);

so many questions given you by Hungarian notation

By bvs23bkv33, 2017-06-09 08:24:49

// enum - full enumeration of knapsack solutions
// (C) Joshua Knowles

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <stdbool.h>

FILE *fp;  // file pointer for reading the input files
int Capacity;     // capacity of the knapsack (total weight that can be stored)
int Nitems;    // number of items available
int *item_weights;  // vector of item weights
int *item_values;  // vector of item profits or values
int *temp_indexes;  // list of temporary item indexes for sorting items by value/weight
int QUIET=0; // this can be set to 1 to suppress output

extern void read_knapsack_instance(char *filename);
extern void print_instance();
extern void sort_by_ratio();
extern int check_evaluate_and_print_sol(int *sol,  int *total_value, int *total_weight);
void enumerate();
int next_binary(int *str, int Nitems);

int main(int argc, char *argv[])
{
  read_knapsack_instance(argv[1]);
  print_instance();
  enumerate();
  return(0);
}

void enumerate()
{
  // Do an exhaustive search (aka enumeration) of all possible ways to pack
  // the knapsack.
  // This is achieved by creating every binary solution vector of length Nitems.
  // For each solution vector, its value and weight is calculated.


  int i;  // item index
  int solution[Nitems+1];   // (binary) solution vector representing items packed
  int best_solution[Nitems+1];  // (binary) solution vector for best solution found
  int best_value; // total value packed in the best solution
  double j=0;
  int total_value, total_weight; // total value and total weight of current knapsack solution
  int infeasible;  // 0 means feasible; -1 means infeasible (violates the capacity constraint)

  // set the knapsack initially empty
  for(i=1;i<=Nitems;i++)
    {
      solution[i]=0;
    }
  QUIET=1;
  best_value=0;

 while(!(next_binary(&solution[1], Nitems)))
    {

      /* ADD CODE IN HERE TO KEEP TRACK OF FRACTION OF ENUMERATION DONE */

          // calculates the value and weight and feasibility:
      infeasible=check_evaluate_and_print_sol(solution, &total_value, &total_weight);  
      /* ADD CODE IN HERE TO KEEP TRACK OF BEST SOLUTION FOUND*/

    }
 /* ADD CODE TO PRINT OUT BEST SOLUTION */

}


int next_binary(int *str, int Nitems)
{
  // Called with a binary string of length Nitems, this 
  // function adds "1" to the string, e.g. 0001 would turn to 0010.
  // If the string overflows, then the function returns 1, else it returns 0.
  int i=Nitems-1;
  while(i>=0)
    {
      if(str[i]==1)
	{
	  str[i]=0;
	  i--;
	}
      else
	{
	  str[i]=1;
	  break;
	}
    }
  if(i==-1)
    {
      return(1);
    }
  else
    {
      return(0);
    }
}

A genuine UoM lab.

By Joshua Knowles, 2018-02-23 10:58:03
int t;

if ((t > 1) && (t < 2))
{
    errorString = errorBuffer;
	return -1;
}
By Anonymous, 2017-12-13 19:42:08
int min(int a,int b,int c) //Function to return the minimum.
{
  if(a < b)
  {
    if(a < c)return a;
    else if(a > c)return c;
    else return a;
  }
  else if(a > b)
  {
    if(b < c)return b;
    else if(b > c)return c;
    else return b;
  }
  else
  {
    if(a < c) return a;
    else if(a > c) return c;
    else return a;
  }
}

This kind of things make me hate my work.

By Another random student of algorithms., 2017-12-14 00:45:52
//shit code 80 lvl
void getMessageAndChannel(char*buffer,char*message,char*channel)
{
if(strstr(buffer, "PRIVMSG") != NULL )
{
while(*buffer !='#')*buffer++;
while(*buffer!=' ' && *buffer)
 *channel++=*buffer++;
while(*buffer!=':' && *buffer)*buffer++;
while(*buffer!='\n' && *buffer)
*message++=*buffer++;
}///
}
By Warlock-Dalbaeb, 2017-03-25 20:54:45
...
if (alignEntrySize & 1)
    alignEntrySize++;
if (alignEntrySize & 2)
    alignEntrySize += 2;
...
By dzuma, 2018-02-08 18:19:17
s[strlen(s)] = '\0';
By Ulfalizer, 2017-12-13 19:17:36
file = fopen(argv[1], "r");

if (file == NULL){}
    exit(EXIT_FAILURE);

Those silly curly braces.

By Anonymous, 2017-12-14 05:47:47
int true = 0;
while (true)
{
    //do something
}

true = false

By noz1995, 2017-12-15 16:08:59