warning: format ‘%f’ expects argument of type ‘float ’, but argument 2 has type ‘double ’

Question

everyone. I NEED HELP! I was trying to submit this following HackerRank's challenge : Task Given the meal price (base cost of a meal), tip percent (the percentage of the meal price being added as tip), and tax percent (the percentage of the meal price being added as tax) for a meal, find and print the meal's total cost. Round the result to the nearest integer.

#include <stdio.h>
#include <math.h>
int main()

{
    int tax,tip;
    double mealc;
    
scanf("%f",&mealc);
scanf("d",&tip);
scanf("%d",&tax);
mealc = mealc+(mealc*tip/100))+(mealc*tax/100);
printf ("%d",round(mealc));

    return 0;
}

After compiling the code above. I always get these errors :

Hk2.c:33:9: warning: format ‘%f’ expects argument of type ‘float *’, but argument 2 has type ‘double *’ [-Wformat=]

Hk2.c:37:11: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘double’ [-Wformat=]

What is the problem ?

Why do you think that `scanf("%f",&mealc)` is correct, and what do you think the warning message is trying to tell you? — Steve Summit, Nov 29 '21 at 21:21
`%f` formats a `float` type, hence the arguments tou need to pass to `scanf` should be a float — Rafaelplayerxd YT, Nov 29 '21 at 21:26
You were right to have used type `double`; it is preferable in most circumstances (rather than using `float`, which is often too imprecise). So favor Wiliam Pursell's comment, and ignore Rafaelplayerxd's and Tzatziki's. — Steve Summit, Nov 29 '21 at 21:33

score 1 · Answer 1 · answered Nov 29 '21 at 21:23

1

Change mealc to float. At your second scanf tou missed a %: scanf("%d",&tip);

answered Nov 29 '21 at 21:23

Tzatziki Xwris Kremmudi

11
1

score 0 · Accepted Answer · answered Nov 29 '21 at 21:34

0

As the warning message says the conversion specifier %f is designated to input values for objects of the type float instead of the type double.

To input a value for an object of the type double you need to use the conversion specifier %lf.

scanf("%lf",&mealc);

Also you have a typo in this call

scanf("d",&tip);

you need to write

scanf("%d",&tip);

And in this statement

mealc = mealc+(mealc*tip/100))+(mealc*tax/100);

there is a redundant closing parenthesis. You need to write

mealc = mealc+(mealc*tip/100)+(mealc*tax/100);

answered Nov 29 '21 at 21:34

Vlad from Moscow

301,070
26
186
335

` #include #include int main() { int tax,tip; double mealc; scanf("%lf",&mealc); scanf("%d",&tip); scanf("%d",&tax); mealc = mealc+(mealc*tip/100)+(mealc*tax/100); printf ("%f",round(mealc)); return 0; } ` Yeah, it worked fine, but the result is float type. When I change the type I get the same error. – Cyclone Nov 29 '21 at 22:02
If you're worried about the last printout containing a decimal point, either use `%.0f`, or maybe `%d` along with `(int)round(mealc)`. – Steve Summit Nov 29 '21 at 22:08
Thanks, it worked.. – Cyclone Nov 29 '21 at 22:14
What is the difference between `"%f"` and `"%lf"`? – Cyclone Nov 29 '21 at 22:20
@Cyclone Used with the function scanf the conversion specifier %f expects a pointer to an object of the type float while the conversion specifier %lf expects a pointer to an object of the type double. – Vlad from Moscow Nov 29 '21 at 22:24

warning: format ‘%f’ expects argument of type ‘float ’, but argument 2 has type ‘double ’

2 Answers2