implicit declatation of function 'int toupper(...)'

WTF does this mean it's pointing to this code...

//Michael Handlon
//Project 4

#include <stdio.h>

int answer = 0;

char num1;
int tmp1 = 0;

char oper;

char num2;
int tmp2 = 0;

int main(void){
printf("Input your first Roman Numeral (1 @ a time; # to quit) ");


while((num1 = getchar()) != '#'){

switch( toupper( num1 ) ){ // says error is on this line.

case 'I':
tmp1 = tmp1 + 1;
break;

case 'V':
tmp1 = tmp1 + 5;
break;

case 'X':
tmp1 = tmp1 + 10;
break;

case 'L':
tmp1 = tmp1 + 50;
break;

case 'C':
tmp1 = tmp1 + 100;
break;

case 'D':
tmp1 = tmp1 + 500;
break;

case 'M':
tmp1 = tmp1 + 1000;
break;

case '#':
printf("Quit then.");
break;

default:
printf("WTF are you doing?");
break;
} // end switch
} // end while

printf("What kind of operator would you like to use('+', '-', '*', '/'?): ");
oper = getchar();


printf("Input your second Roman Numeral (1 @ a time; # to quit) ");

while((num2 = getchar()) != '#'){

switch( toupper( num2 ) ){

case 'I':
tmp2 = tmp2 + 1;
break;

case 'V':
tmp2 = tmp2 + 5;
break;

case 'X':
tmp2 = tmp2 + 10;
break;

case 'L':
tmp2 = tmp2 + 50;
break;

case 'C':
tmp2 = tmp2 + 100;
break;

case 'D':
tmp2 = tmp2 + 500;
break;

case 'M':
tmp2 = tmp2 + 1000;
break;

case '#':
printf("Quit then.");
break;

default:
printf("WTF are you doing?");
break;
} //end switch
} //end while


if (oper == '+'){
answer = tmp1 + tmp2;
}

if (oper == '-'){
answer = tmp1 - tmp2;
}

if (oper == '*'){
answer = tmp1 * tmp2;
}

if (oper == '/'){
answer = tmp1 / tmp2;
}

printf("Your answer is %d.",answer);

return 0;
}

//Michael Handlon
//Project 4

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

int answer = 0;

char num1;
int tmp1 = 0;

char oper;

char num2;
int tmp2 = 0;

int main(void){
printf("\nInput your first Roman Numeral (1 @ a time; # to quit) ");
while((num1 = getchar()) != '#'){


switch( toupper( num1 ) ){

case 'I':
tmp1 = tmp1 + 1;
printf("\nInput your first Roman Numeral (1 @ a time; # to quit) ");
break;

case 'V':
tmp1 = tmp1 + 5;
printf("\nInput your first Roman Numeral (1 @ a time; # to quit) ");
break;

case 'X':
tmp1 = tmp1 + 10;
printf("\nInput your first Roman Numeral (1 @ a time; # to quit) ");
break;

case 'L':
tmp1 = tmp1 + 50;
printf("\nInput your first Roman Numeral (1 @ a time; # to quit) ");
break;

case 'C':
tmp1 = tmp1 + 100;
printf("\nInput your first Roman Numeral (1 @ a time; # to quit) ");
break;

case 'D':
tmp1 = tmp1 + 500;
printf("\nInput your first Roman Numeral (1 @ a time; # to quit) ");
break;

case 'M':
tmp1 = tmp1 + 1000;
printf("\nInput your first Roman Numeral (1 @ a time; # to quit) ");
break;

case '#':
printf("\nQuit then.");
break;

} // end switch
} // end while

printf("\nWhat kind of operator would you like to use('+', '-', '*', '/'?): ");
oper = getchar();

printf("\nInput your second Roman Numeral (1 @ a time; # to quit) ");
while((num2 = getchar()) != '#'){

switch( toupper( num2 ) ){

case 'I':
tmp2 = tmp2 + 1;
printf("\nInput your second Roman Numeral (1 @ a time; # to quit) ");
break;

case 'V':
tmp2 = tmp2 + 5;
printf("\nInput your second Roman Numeral (1 @ a time; # to quit) ");
break;

case 'X':
tmp2 = tmp2 + 10;
printf("\nInput your second Roman Numeral (1 @ a time; # to quit) ");
break;

case 'L':
tmp2 = tmp2 + 50;
printf("\nInput your second Roman Numeral (1 @ a time; # to quit) ");
break;

case 'C':
tmp2 = tmp2 + 100;
printf("\nInput your second Roman Numeral (1 @ a time; # to quit) ");
break;

case 'D':
tmp2 = tmp2 + 500;
printf("\nInput your second Roman Numeral (1 @ a time; # to quit) ");
break;

case 'M':
tmp2 = tmp2 + 1000;
printf("\nInput your second Roman Numeral (1 @ a time; # to quit) ");
break;

case '#':
printf("\nQuit then.");
printf("\nInput your second Roman Numeral (1 @ a time; # to quit) ");
break;

} //end switch
} //end while


if (oper == '+'){
answer = tmp1 + tmp2;
}

if (oper == '-'){
answer = tmp1 - tmp2;
}

if (oper == '*'){
answer = tmp1 * tmp2;
}

if (oper == '/'){
answer = tmp1 / tmp2;
}

printf("\nYour answer is %d.",answer);

return 0;
}



Any idea why this compiles yet hangs after you add an operator?

As for your char oper; instead of using " + - * / " say something like.....this compiles and runs...

Code:

printf("Enter mode of operation"
          " A for add, S for subtract"
          "M for multiply, D for divide");
oper = toupper(getch());

switch (oper) {
      case A: /*now perform add */
             break;
      case S: /*perform subtract */
      /*same with multiply and divide  */

I have compiled your program using my way and it compiles and runs....minor alterations....user inputs operation after second roman numerals....and no need for variable char oper.
pm me for full code

too long for pm...here it is then.....

Code:

#include <stdio.h> 
#include <ctype.h>
#include <conio.h>  /*for getch() fn  */

int main(void)
{
int answer = 0; 
char num1;
int tmp1 = 0;
char num2;
int tmp2 = 0;

printf("\nInput your first Roman Numeral (1 @ a time; # to quit) "); 

/*NOTICE use of getch as defined in conio.h*/

while((num1 = getch()) != '#'){ 


switch( toupper( num1 ) ){ 

case 'I': 
tmp1 = tmp1 + 1; 
printf("\nInput your first Roman Numeral (1 @ a time; # to quit) "); 
break; 

case 'V': 
tmp1 = tmp1 + 5; 
printf("\nInput your first Roman Numeral (1 @ a time; # to quit) "); 
break; 

case 'X': 
tmp1 = tmp1 + 10; 
printf("\nInput your first Roman Numeral (1 @ a time; # to quit) "); 
break; 

case 'L': 
tmp1 = tmp1 + 50; 
printf("\nInput your first Roman Numeral (1 @ a time; # to quit) "); 
break; 

case 'C': 
tmp1 = tmp1 + 100; 
printf("\nInput your first Roman Numeral (1 @ a time; # to quit) "); 
break; 

case 'D': 
tmp1 = tmp1 + 500; 
printf("\nInput your first Roman Numeral (1 @ a time; # to quit) "); 
break; 

case 'M': 
tmp1 = tmp1 + 1000; 
printf("\nInput your first Roman Numeral (1 @ a time; # to quit) "); 
break; 

case '#': 
printf("\nQuit then."); 
break; 

} // end switch 
} // end while

/* NOTICE    */
/*input of operation moved to after second roman input*/

printf("\nInput your second Roman Numeral (1 @ a time; # to quit) "); 

/*NOTICE use of getch as defined in conio.h*/

while((num2 = getch()) != '#'){ 

switch( toupper( num2 ) ){ 

case 'I': 
tmp2 = tmp2 + 1; 
printf("\nInput your second Roman Numeral (1 @ a time; # to quit) "); 
break; 

case 'V': 
tmp2 = tmp2 + 5; 
printf("\nInput your second Roman Numeral (1 @ a time; # to quit) "); 
break; 

case 'X': 
tmp2 = tmp2 + 10; 
printf("\nInput your second Roman Numeral (1 @ a time; # to quit) "); 
break; 

case 'L': 
tmp2 = tmp2 + 50; 
printf("\nInput your second Roman Numeral (1 @ a time; # to quit) "); 
break; 

case 'C': 
tmp2 = tmp2 + 100; 
printf("\nInput your second Roman Numeral (1 @ a time; # to quit) "); 
break; 

case 'D': 
tmp2 = tmp2 + 500; 
printf("\nInput your second Roman Numeral (1 @ a time; # to quit) "); 
break; 

case 'M': 
tmp2 = tmp2 + 1000; 
printf("\nInput your second Roman Numeral (1 @ a time; # to quit) "); 
break; 

case '#': 
printf("\nQuit then."); 
printf("\nInput your second Roman Numeral (1 @ a time; # to quit) "); 
break; 

} //end switch 
} //end while 

/*NOTICE  HERE is operation input  */

printf("Enter mode of operation"
          " A for add, S for subtract"
         "M for multiply, D for divide");

/*NO NEED FOR oper TOUPPER AND GETCH fn take care*/
/*of user input*/
  switch(toupper(getch()))
  {
    case 'A': answer = tmp1 + tmp2;/*now perform add */
             break;
    case 'S': answer = tmp1 - tmp2;/*perform subtract */
             break;
    case 'M': answer = tmp1 * tmp2;/*now multiply*/
             break;
    case 'D': answer = tmp1 / tmp2; /*finally divide*/
   }
printf("\nYour answer is %d.",answer);

return 0; 
}