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Width format conversion
In the following code..the output should be: "New price is 8.50" but the decimal part in the number (8.50) which is (8) should be having 4 digits reserved for it..but it doesn't and I really don't understand it...can anyone please explain this to me ? ------------------------------------ #include <stdio.h> int main() { float price = 6.50; int increase = 2; float new_price; new_price = price + increase; printf("New price is %4.2f", new_price); /* Output: New price is 8.50 */ return 0; }
4 ответов
+ 4
/*Width format conversion
In the following code..the output should be: "New price is 8.50" but the decimal part in the number (8.50) which is (8) should be having 4 digits reserved for it..but it doesn't and I really don't understand it...can anyone please explain this to me ?
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#include <stdio.h>
int main() {
float price = 6.50;
int increase = 2;
float new_price;
new_price = price + increase;
printf("%06.2f", new_price); // with 0 as fillers
/* Output: New price is 8.50 */
printf("\n") ;
printf("%6.2f", new_price); // with Space as fillers
return 0;
}
+ 3
Run the above the code and see the difference
(The width field is for the entire converted string not just the whole number part).
+ 3
Mohamed Taha Yes!
+ 1
@RKK
Run the above the code and see the difference
(The width field is for the entire converted string not just the whole number part).
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So this means the decimal point is also counted in the 6 reserved digits ?