hello all,
why the output is always zero when i try to convert it from int to float in the printf function. till i explicitly cast it to a float it keeps on outping zeros. here is the code
and my outputCode:#include<stdio.h> int main() { int a=10; //int *p=&a; printf("%f",a); getchar(); }
can anyone tell me why this stange output isCode:0.000000
thank u
-onthewaytoC
Because you're not converting it -- you're just lying to printf and seeing some form of undefined behavior. Either tell the truth and use %d, or perform an actual conversion.Code:#include <stdio.h> int main() { int a = 10; printf("%d\n", a); printf("%f\n", (float)a); getchar(); return 0; } /* my output 10 10.000000 */
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
Try this instead:
Code:printf("%f", (float) a);
Or if you prefer:
Takes less keystrokes.Code:printf("%f", 0.0f + a );
Quzah.
Hope is the first step on the road to disappointment.
thax very for your help, i have one more question to be clarified. some text books still follows conversion of type in this method (especially) converting from int to char using printf
is this a proper way of converting. the question remains the same, cose it converts int to char using printf but not int to floatCode:#include<stdio.h> int main() { int a=65; printf("%c",a); getchar(); return 0; }
-onthewaytoC
Well certainly in ASCII, 65 is equal to "A", therefore the int already "is" the character. All the printf statement does is treat the int like a character instead of an integer, so it prints a character instead of a number.
