Double problem

Hello again. I have problem with double printing.

Code:

---

#include <stdio.h>

int main()
{
 double a=0;

 printf("Give a Float number (ex. 346.58)");
 scanf("%e", &a);

 printf("\n\n---->%f\n\n", a);
 
return 0;
}

---

I run:

Give a Float number (ex. 346.58)346.58


---->0.000000

Why this and how I can correct it. I want double but I don't want that format 5.609771e-315.

Thank you again.

%e is used for normal sized floating point numbers (float). To enter a double precision number, you need to add a modifier. So it should look like "%le".

Quote:

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Originally Posted by bithub

%e is used for normal sized floating point numbers (float). To enter a double precision number, you need to add a modifier. So it should look like "%le".

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With %le the result is 5.609771e-315.


But I want 346.58.

you also have to change the printf code too

printf and scanf format codes

Quote:

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Originally Posted by Epy

you also have to change the printf code too

printf and scanf format codes

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No, printf is not the same as scanf. %f is fine for printing doubles.

I'm not sure what the problem is cs05pp2. I think your code should work with the change to scanf(). What compiler/OS are you using?

Quote:

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Originally Posted by bithub

No, printf is not the same as scanf. %f is fine for printing doubles.

I'm not sure what the problem is cs05pp2. I think your code should work with the change to scanf(). What compiler/OS are you using?

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I use two kind: FreeBSD and Fedora

Quote:

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Originally Posted by cs05pp2

I use two kind: FreeBSD and Fedora

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Neither of those are compilers.

Quote:

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Originally Posted by bithub

Neither of those are compilers.

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GCC Compiler.

I think the actual problem is that %e is used for floating point numbers in exponential form. So if you wanted to input a double the way you're trying now, you should use %lf. Otherwise, with %le, you'd need to type "3.4658e+02"

Quote:

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Originally Posted by Ronix

I think the actual problem is that %e is used for floating point numbers in exponential form. So if you wanted to input a double the way you're trying now, you should use %lf. Otherwise, with %le, you'd need to type "3.4658e+02"

---

Code:

---

#include <stdio.h>

int main()
{
 double a=0;
 printf("Give a Float number (ex. 346.58)");
 scanf("%e", &a);

 printf("\n\n---->%lf\n\n", a);
 
return 0;
}

---

I try that and I get :

Code:

---

Give a Float number (ex. 346.58)346.58


---->0.000000

---

Right, because:

- You're using %e when you should be using %le
- Your input is not in exponential form. Type "3.4658e+02" and see if it works.

Otherwise, change the %e to %lf and keep the input as is.

Quote:

---

Originally Posted by Ronix

Right, because:

- You're using %e when you should be using %le
- Your input is not in exponential form. Type "3.4658e+02" and see if it works.

Otherwise, change the %e to %lf and keep the input as is.

---

Perfect thank you