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c++scanfformat-specifiers
Ahmed
Ahmed

Reputation: 1359

What is the problem with the format specifier C++

I have the following code:

#include <cstdio>

int main()
{
    float b = 323.23f;
    std::scanf("%6.3f\n", &b);  // <-- Warning
    std::printf("%6.3f\n", b);
}

There is a warning at scanf() stating:

Invalid conversion specifier '.'

Is there something I am missing here?

Upvotes: 2

Views: 417

Answers (3)

rob mayoff
rob mayoff

Reputation: 386038

When you use %6.3f with printf, the .3 specifies the precision of the output.

scanf doesn't need you to specify the precision, because it will accept whatever precision it finds in the input stream. So you aren't allowed to specify .3 (or .-anything-else) in a scanf format. Just use %6f or even %f instead.

Upvotes: 8

AsH
AsH

Reputation: 31

The format specifier in C does not take the value '6.3'.You can use '%6f' or simply '%f'. '%6.3f' is the thing that is causing the warning.

Upvotes: 1

Harshith Rai
Harshith Rai

Reputation: 3066

The format specifier for scanf doesn't require you to explicitly specify the number of places after the decimal point. Plus, it is what is causing the warning, it is better off to use just a plain %f inside the scanf though a %6f also works.

This isn't the case with the printf though. When you use a [some_number].[number_of_places]f, like, for example, 6.3f in a printf, while the .3f would affect the number of places after the decimal point which would be displayed, the 6 in it reserves that many character spaces to display the entire number.

Hope this helps to some extent.

Upvotes: 1

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