Programming Forums > Application Development > C

linuxpimp20 · Jul 5th, 2005, 4:38 PM

Im having trouble with an exercise in the C book written by the inventors of C. Normally i don't like people who post code and ask what is wrong with it but i've been working on this since this morning with no success. I don't have much choice because the C book doesn't have the answers in the back of the book. Im not doign this for a class im interesting in learning C and that is why i want to know what im doing wrong. Hoping someone can look at it because the more i look at it the more it just blends together and i can't pick out what is wrong. In a previous exercise i wrote a program that converts fahrneheit to celsius and displays the conversions from 0 to 300 in steps of 20. Now it is asking for me to have it display the output going from 300 to 0. this is what i have:

main()
{
int fahr;

printf("Fahr\t Celsius\n");
for (fahr = 300; fahr >= 0; fahr = fahr - 20);
printf("%3.0f \t%6.1f\n", fahr, (5.0/9.0) * (fahr-32.0));

}

here is the output i get:

Fahr Celsius
0 0.0

i wouldn't have posted if this thing hasn't frustrated me so much because i know it is just something simple im overlooking. thanks for any replies in advance.

EverLearning · Jul 5th, 2005, 4:52 PM

(Toggle Plain Text)

main()
{
    int fahr;

    printf("Fahr\t Celsius\n");
    for (fahr = 300; fahr >= 0; fahr = fahr - 20);
      printf("%3.0f \t%6.1f\n", fahr, (5.0/9.0) * (fahr-32.0));

}

For starters, get rid of semicolumn, since you want to have printf inside the loop.
I would always put {} around loops and other control structures, although in this case it does not matter (since it's just one statement).

stevengs · Jul 5th, 2005, 5:02 PM

how about something like this:

(Toggle Plain Text)

#include <stdio.h>


int main(int argc, char* argv[])
{
     int fahr;

     printf("Fahr\t Celsius\n");
     for (fahr = 300; fahr >= 0; fahr = fahr - 20)
          printf("%i \t %6.1f\n", fahr,  (5.0/9.0) * (fahr-32.0));

     return 0;
}

-EDIT
(double check what is typed in the book! I would not try to cast an int to FLOAT implicitly by passing it to the printf function... if necessary, do an explicit cast with (FLOAT) ) :

(Toggle Plain Text)

printf("%3.0f \t %6.1f\n", (float)fahr,  (5.0/9.0) * (fahr-32.0));

EverLearning · Jul 5th, 2005, 5:17 PM

being picky

it would be a float.
EDIT: if in printf() you use %f that is.
Oh, whatever....

stevengs · Jul 5th, 2005, 5:25 PM

right... I beg your pardon - %lf must be double then... ( unless you are using a nonconforming M$

, which I have for waaay to long... (4 months, hehe ))

EverLearning · Jul 5th, 2005, 5:32 PM

Actually I think you can do it with %f as well, my bet here.
This I found helpful before (...'coz I often forget this stuff!), took several minutes to find it on google: table
So, the only difference between float and double is the size then.

linuxpimp20 · Jul 6th, 2005, 7:22 AM

thank you that was my problem that i needed to cast fahr. that initially made me feel like something was arwy with the int and the printf being %f. I've spent most of my time learning about C++ so i still need to get used to printf function where you have to specific everything. sorry for not posting sooner but the post weren't getting sent to me via email.

Jul 5th, 2005, 4:38 PM	#1
linuxpimp20 Hobbyist Programmer Join Date: May 2005 Location: ma Posts: 130 Rep Power: 4	Having trouble with an exercise in the C book. Im having trouble with an exercise in the C book written by the inventors of C. Normally i don't like people who post code and ask what is wrong with it but i've been working on this since this morning with no success. I don't have much choice because the C book doesn't have the answers in the back of the book. Im not doign this for a class im interesting in learning C and that is why i want to know what im doing wrong. Hoping someone can look at it because the more i look at it the more it just blends together and i can't pick out what is wrong. In a previous exercise i wrote a program that converts fahrneheit to celsius and displays the conversions from 0 to 300 in steps of 20. Now it is asking for me to have it display the output going from 300 to 0. this is what i have: main() { int fahr; printf("Fahr\t Celsius\n"); for (fahr = 300; fahr >= 0; fahr = fahr - 20); printf("%3.0f \t%6.1f\n", fahr, (5.0/9.0) * (fahr-32.0)); } here is the output i get: Fahr Celsius 0 0.0 i wouldn't have posted if this thing hasn't frustrated me so much because i know it is just something simple im overlooking. thanks for any replies in advance.

Jul 5th, 2005, 4:52 PM	#2
EverLearning Hobbyist Programmer Join Date: May 2005 Location: Indiana Posts: 130 Rep Power: 4	(Toggle Plain Text) main() { int fahr; printf("Fahr\t Celsius\n"); for (fahr = 300; fahr >= 0; fahr = fahr - 20); printf("%3.0f \t%6.1f\n", fahr, (5.0/9.0) * (fahr-32.0)); } main() { int fahr; printf("Fahr\t Celsius\n"); for (fahr = 300; fahr >= 0; fahr = fahr - 20); printf("%3.0f \t%6.1f\n", fahr, (5.0/9.0) * (fahr-32.0)); } For starters, get rid of semicolumn, since you want to have printf inside the loop. I would always put {} around loops and other control structures, although in this case it does not matter (since it's just one statement). __________________ got math? yumm... "All men by nature desire to know" -Aristotle, Metaphysics Last edited by EverLearning; Jul 5th, 2005 at 5:11 PM.

Jul 5th, 2005, 5:02 PM	#3
stevengs Professional Programmer Join Date: May 2005 Location: Bad Nauheim, Germany Posts: 436 Rep Power: 4	how about something like this: (Toggle Plain Text) #include <stdio.h> int main(int argc, char* argv[]) { int fahr; printf("Fahr\t Celsius\n"); for (fahr = 300; fahr >= 0; fahr = fahr - 20) printf("%i \t %6.1f\n", fahr, (5.0/9.0) * (fahr-32.0)); return 0; } #include <stdio.h> int main(int argc, char* argv[]) { int fahr; printf("Fahr\t Celsius\n"); for (fahr = 300; fahr >= 0; fahr = fahr - 20) printf("%i \t %6.1f\n", fahr, (5.0/9.0) * (fahr-32.0)); return 0; } -EDIT (double check what is typed in the book! I would not try to cast an int to FLOAT implicitly by passing it to the printf function... if necessary, do an explicit cast with (FLOAT) ) : (Toggle Plain Text) printf("%3.0f \t %6.1f\n", (float)fahr, (5.0/9.0) * (fahr-32.0)); printf("%3.0f \t %6.1f\n", (float)fahr, (5.0/9.0) * (fahr-32.0)); __________________ -Steven "Is this a piece of your brain?" - Basil Fawlty Last edited by stevengs; Jul 5th, 2005 at 5:27 PM.

Jul 5th, 2005, 5:17 PM	#4
EverLearning Hobbyist Programmer Join Date: May 2005 Location: Indiana Posts: 130 Rep Power: 4	being picky it would be a float. EDIT: if in printf() you use %f that is. Oh, whatever.... __________________ got math? yumm... "All men by nature desire to know" -Aristotle, Metaphysics

Jul 5th, 2005, 5:25 PM	#5
stevengs Professional Programmer Join Date: May 2005 Location: Bad Nauheim, Germany Posts: 436 Rep Power: 4	right... I beg your pardon - %lf must be double then... ( unless you are using a nonconforming M$ , which I have for waaay to long... (4 months, hehe )) __________________ -Steven "Is this a piece of your brain?" - Basil Fawlty Last edited by stevengs; Jul 5th, 2005 at 5:38 PM.

Jul 5th, 2005, 5:32 PM	#6
EverLearning Hobbyist Programmer Join Date: May 2005 Location: Indiana Posts: 130 Rep Power: 4	Actually I think you can do it with %f as well, my bet here. This I found helpful before (...'coz I often forget this stuff!), took several minutes to find it on google: table So, the only difference between float and double is the size then. __________________ got math? yumm... "All men by nature desire to know" -Aristotle, Metaphysics

Jul 6th, 2005, 7:22 AM	#7
linuxpimp20 Hobbyist Programmer Join Date: May 2005 Location: ma Posts: 130 Rep Power: 4	thank you that was my problem that i needed to cast fahr. that initially made me feel like something was arwy with the int and the printf being %f. I've spent most of my time learning about C++ so i still need to get used to printf function where you have to specific everything. sorry for not posting sooner but the post weren't getting sent to me via email.

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Programming Forums > Application Development > C

Having trouble with an exercise in the C book.