Programming Forums > Application Development > C

sparda · Oct 25th, 2005, 8:59 PM

Im doing a simple Convertion program. It converts from Fahrenheit to Celsius and Visa Versa. My problem involves the scanf Function. Im really curious why it happens; the function has trouble handling Char types.

for example, code exerpt:

(Toggle Plain Text)

char choice;
printf("\n\t--If converting from Celsius to Fahrenheit, please press 'C'-");
printf("\n\t--If converting from Fahrenheit to Celsius, please press 'F'-\n");
     printf("\n\tChoice:");
     scanf("%c", &choice);

The Code above does not work, i mean it compiles, but it doesnt store a value at 'choice'.

but strangely, the code below works, and its almost exactly the same ( the only difference is that the %c is replaced with %s to handle strings, but 'choice' is not a string!

(Toggle Plain Text)

char choice;
printf("\n\t--If converting from Celsius to Fahrenheit, please press 'C'-");
printf("\n\t--If converting from Fahrenheit to Celsius, please press 'F'-\n");
     printf("\n\tChoice:");
     scanf("%s", &choice);

Im using Dev C++, and consider myself an intermidiate C programmer, so please dont tell me to go learn something easier.
Im guessing it has something to do with how the function was written, but i cant find the actual code on stdio.h or stdlib.h .
Can someone please tell me why this happens? any help will be appreciated, thanks.

nindoja · Oct 25th, 2005, 11:44 PM

Here is a suggestion, post the rest of the code, because the input stream might be catching an extra enter, and it is hard to tell with the code we have. Also, the scanf with the %s should NOT have an ampersand.

Benoit · Oct 25th, 2005, 11:53 PM

Your first piece of code should work...Can you post your entire program?

Kaja Fumei · Oct 26th, 2005, 12:45 AM

Quote:

Originally Posted by nindoja

Also, the scanf with the %s should NOT have an ampersand.

Technically it should, because the variable being stored in is still a char so you need to pass its pointer (However, I do not recommend leaving the variable as a char when using %s).

Polyphemus_ · Oct 26th, 2005, 7:48 AM

Quote:

Originally Posted by Kaja Fumei

Technically it should, because the variable being stored in is still a char so you need to pass its pointer (However, I do not recommend leaving the variable as a char when using %s).

When using a char array, you always use its address. Using an ampersant is in many compilers still correct, because they remove it - it's a common mistake.

Benoit · Oct 26th, 2005, 7:49 AM

You don't have to use the & because it alrady holds an address no?

grumpy · Oct 26th, 2005, 8:00 AM

Sort of correct, Benoit. The name of a char array is not the same as it's address, but is treated by the compiler in some contexts as if it is

sparda · Oct 26th, 2005, 2:54 PM

Here is the entire code, as nindoja said, i think it has something to do
with the input stream.

(Toggle Plain Text)

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



int main()
{
     
     float cels, farh, var1;
     char choice;
     // Intro 
          printf("\t\tThis is a program that will convert\n");
          printf("\t\ta temperature from Fahrenheit to Celsius or\n");
          printf("\t\tfrom Celsius to Fahrenheit.\n\n\n");


     
     // 
     printf("\t\tPlease Select a Temperature: ");
     scanf("%f", &var1);
     cels = (5.0/9.0)*(var1 - 32);
     farh = (9.0/5.0)*(var1 + 32);

     printf("\n\t--If converting from Celsius to Fahrenheit, please press 'C'--");
     printf("\n\t--If converting from Fahrenheit to Celsius, please press 'F'--\n");
     printf("\n\tChoice:");
     scanf("%s", &choice);
     
     if(choice == 'C' || choice == 'c') 
     {
          printf("\n\t%5.2f degrees Celsius is converted to %5.2f degrees Fahrenheit\n", var1, farh);
          system("pause");
          
     }
          
     if(choice == 'F' || choice == 'f')  
     {
          printf("\n\t%5.2f degrees Fahrenheit is converted to %5.2f degrees Celsius\n", var1, cels);
          system("pause");
          
     }
          
     system("pause");

     
     
       
     return 0;
}

jim mcnamara · Oct 26th, 2005, 3:27 PM

This works in terms of I/O and getting variables read in correctly.
The formulas are implemented incorrectly - you get wrong answers - you can fix that.
I added a switch() -by "standard" convention whenever you have three or more condtions

(choice = C or c or F or c and choice not = "F,f,C,c" ---> five tests)

you employ a switch().

(Toggle Plain Text)

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

int main()
{
     double cels=0.,
            farh=0.,
            var1=0.;
     char choice[10]={0x0};

     printf("\t\tThis is a program that will convert\n");
     printf("\t\ta temperature from Fahrenheit to Celsius or\n");
     printf("\t\tfrom Celsius to Fahrenheit.\n\n\n");
     printf("\t\tPlease Select a Temperature: ");
     scanf("%s", choice);
     var1=atof(choice);
     /* the math here is WRONG */
     cels = ((5.0/9.0)*var1) - 32.;
     farh = ((9.0/5.0)*var1) + 32.;
     printf("\n\t--If converting from Celsius to Fahrenheit, please press 'C'<return>--\n");
     printf("\n\t--If converting from Fahrenheit to Celsius, please press 'F'<return>--\n");
     printf("\n\tChoice:\n");
     scanf("%s", &choice);
     switch(choice[0])
     {
            case 'C':
            case 'c':
                printf("\n\t%5.2f degrees Celsius is converted to %5.2f degrees Fahrenheit\n",
                   var1, farh);
                break;
            case 'F':
            case 'f':                 
               printf("\n\t%5.2f degrees Fahrenheit is converted to %5.2f degrees Celsius\n", 
                  var1, cels);                
            default :     
                break;
     }
     system("pause");
     return 0;
}

Oct 25th, 2005, 8:59 PM	#1
sparda Newbie Join Date: Apr 2005 Location: US Posts: 15 Rep Power: 0	problem with standard function Im doing a simple Convertion program. It converts from Fahrenheit to Celsius and Visa Versa. My problem involves the scanf Function. Im really curious why it happens; the function has trouble handling Char types. for example, code exerpt: (Toggle Plain Text) char choice; printf("\n\t--If converting from Celsius to Fahrenheit, please press 'C'-"); printf("\n\t--If converting from Fahrenheit to Celsius, please press 'F'-\n"); printf("\n\tChoice:"); scanf("%c", &choice); char choice; printf("\n\t--If converting from Celsius to Fahrenheit, please press 'C'-"); printf("\n\t--If converting from Fahrenheit to Celsius, please press 'F'-\n"); printf("\n\tChoice:"); scanf("%c", &choice); The Code above does not work, i mean it compiles, but it doesnt store a value at 'choice'. but strangely, the code below works, and its almost exactly the same ( the only difference is that the %c is replaced with %s to handle strings, but 'choice' is not a string! (Toggle Plain Text) char choice; printf("\n\t--If converting from Celsius to Fahrenheit, please press 'C'-"); printf("\n\t--If converting from Fahrenheit to Celsius, please press 'F'-\n"); printf("\n\tChoice:"); scanf("%s", &choice); char choice; printf("\n\t--If converting from Celsius to Fahrenheit, please press 'C'-"); printf("\n\t--If converting from Fahrenheit to Celsius, please press 'F'-\n"); printf("\n\tChoice:"); scanf("%s", &choice); Im using Dev C++, and consider myself an intermidiate C programmer, so please dont tell me to go learn something easier. Im guessing it has something to do with how the function was written, but i cant find the actual code on stdio.h or stdlib.h . Can someone please tell me why this happens? any help will be appreciated, thanks.

Oct 25th, 2005, 11:53 PM	#3
Benoit Expert Programmer Join Date: Sep 2004 Location: Ontario, Canada Posts: 579 Rep Power: 5	Your first piece of code should work...Can you post your entire program? __________________ Johnny was a chemist's son but Johnny is no more, for what Johnny thought was H2O was H2SO4

Oct 26th, 2005, 7:49 AM	#6
Benoit Expert Programmer Join Date: Sep 2004 Location: Ontario, Canada Posts: 579 Rep Power: 5	You don't have to use the & because it alrady holds an address no? __________________ Johnny was a chemist's son but Johnny is no more, for what Johnny thought was H2O was H2SO4

Oct 26th, 2005, 2:54 PM	#8
sparda Newbie Join Date: Apr 2005 Location: US Posts: 15 Rep Power: 0	Here is the entire code, as nindoja said, i think it has something to do with the input stream. (Toggle Plain Text) #include <stdio.h> #include <stdlib.h> int main() { float cels, farh, var1; char choice; // Intro printf("\t\tThis is a program that will convert\n"); printf("\t\ta temperature from Fahrenheit to Celsius or\n"); printf("\t\tfrom Celsius to Fahrenheit.\n\n\n"); // printf("\t\tPlease Select a Temperature: "); scanf("%f", &var1); cels = (5.0/9.0)(var1 - 32); farh = (9.0/5.0)(var1 + 32); printf("\n\t--If converting from Celsius to Fahrenheit, please press 'C'--"); printf("\n\t--If converting from Fahrenheit to Celsius, please press 'F'--\n"); printf("\n\tChoice:"); scanf("%s", &choice); if(choice == 'C' \|\| choice == 'c') { printf("\n\t%5.2f degrees Celsius is converted to %5.2f degrees Fahrenheit\n", var1, farh); system("pause"); } if(choice == 'F' \|\| choice == 'f') { printf("\n\t%5.2f degrees Fahrenheit is converted to %5.2f degrees Celsius\n", var1, cels); system("pause"); } system("pause"); return 0; } #include <stdio.h> #include <stdlib.h> int main() { float cels, farh, var1; char choice; // Intro printf("\t\tThis is a program that will convert\n"); printf("\t\ta temperature from Fahrenheit to Celsius or\n"); printf("\t\tfrom Celsius to Fahrenheit.\n\n\n"); // printf("\t\tPlease Select a Temperature: "); scanf("%f", &var1); cels = (5.0/9.0)(var1 - 32); farh = (9.0/5.0)(var1 + 32); printf("\n\t--If converting from Celsius to Fahrenheit, please press 'C'--"); printf("\n\t--If converting from Fahrenheit to Celsius, please press 'F'--\n"); printf("\n\tChoice:"); scanf("%s", &choice); if(choice == 'C' \|\| choice == 'c') { printf("\n\t%5.2f degrees Celsius is converted to %5.2f degrees Fahrenheit\n", var1, farh); system("pause"); } if(choice == 'F' \|\| choice == 'f') { printf("\n\t%5.2f degrees Fahrenheit is converted to %5.2f degrees Celsius\n", var1, cels); system("pause"); } system("pause"); return 0; }

Oct 26th, 2005, 3:27 PM	#9
jim mcnamara Hobbyist Programmer Join Date: Jun 2005 Location: New Mexico Posts: 228 Rep Power: 4	This works in terms of I/O and getting variables read in correctly. The formulas are implemented incorrectly - you get wrong answers - you can fix that. I added a switch() -by "standard" convention whenever you have three or more condtions (choice = C or c or F or c and choice not = "F,f,C,c" ---> five tests) you employ a switch(). (Toggle Plain Text) #include <stdio.h> #include <stdlib.h> int main() { double cels=0., farh=0., var1=0.; char choice[10]={0x0}; printf("\t\tThis is a program that will convert\n"); printf("\t\ta temperature from Fahrenheit to Celsius or\n"); printf("\t\tfrom Celsius to Fahrenheit.\n\n\n"); printf("\t\tPlease Select a Temperature: "); scanf("%s", choice); var1=atof(choice); /* the math here is WRONG / cels = ((5.0/9.0)var1) - 32.; farh = ((9.0/5.0)var1) + 32.; printf("\n\t--If converting from Celsius to Fahrenheit, please press 'C'<return>--\n"); printf("\n\t--If converting from Fahrenheit to Celsius, please press 'F'<return>--\n"); printf("\n\tChoice:\n"); scanf("%s", &choice); switch(choice[0]) { case 'C': case 'c': printf("\n\t%5.2f degrees Celsius is converted to %5.2f degrees Fahrenheit\n", var1, farh); break; case 'F': case 'f': printf("\n\t%5.2f degrees Fahrenheit is converted to %5.2f degrees Celsius\n", var1, cels); default : break; } system("pause"); return 0; } #include <stdio.h> #include <stdlib.h> int main() { double cels=0., farh=0., var1=0.; char choice[10]={0x0}; printf("\t\tThis is a program that will convert\n"); printf("\t\ta temperature from Fahrenheit to Celsius or\n"); printf("\t\tfrom Celsius to Fahrenheit.\n\n\n"); printf("\t\tPlease Select a Temperature: "); scanf("%s", choice); var1=atof(choice); / the math here is WRONG / cels = ((5.0/9.0)var1) - 32.; farh = ((9.0/5.0)*var1) + 32.; printf("\n\t--If converting from Celsius to Fahrenheit, please press 'C'<return>--\n"); printf("\n\t--If converting from Fahrenheit to Celsius, please press 'F'<return>--\n"); printf("\n\tChoice:\n"); scanf("%s", &choice); switch(choice[0]) { case 'C': case 'c': printf("\n\t%5.2f degrees Celsius is converted to %5.2f degrees Fahrenheit\n", var1, farh); break; case 'F': case 'f': printf("\n\t%5.2f degrees Fahrenheit is converted to %5.2f degrees Celsius\n", var1, cels); default : break; } system("pause"); return 0; }

Oct 25th, 2005, 11:44 PM	#2
nindoja Programmer Join Date: Jun 2005 Posts: 92 Rep Power: 4	Here is a suggestion, post the rest of the code, because the input stream might be catching an extra enter, and it is hard to tell with the code we have. Also, the scanf with the %s should NOT have an ampersand.

Oct 26th, 2005, 8:00 AM	#7
grumpy Programming Guru Join Date: Jun 2005 Location: Adelaide, South Australia Posts: 1,260 Rep Power: 5	Sort of correct, Benoit. The name of a char array is not the same as it's address, but is treated by the compiler in some contexts as if it is

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Programming Forums > Application Development > C

problem with standard function