C++: What is the printf()
format spec for float
? (Visual C++)
It used to be that I used %g
for float
and %lg
for double
.
It looks like the spec changed and float
is undefined and double
is %g
.
I have bits in memory that I am printing out so casting is not an option.
Is there a way that I can print out float
values using printf()
?
Update:
This code was written for unit testing generic C++ libs used on an embedded system.
Here's what I had to do to get the float
to work.
The code is in a template function:
template <typename T,typename TTYP,typename Ttyp,int bits,bool IsSigned>
Error testMatrixT()
{ ...
Here is a code snip:
if (typeid(Ttyp) == typeid(float)) {
float64 c = *(float32*)&Tp(row,col);
float64 a1 = *(float32*)&Arg1(row,col);
float64 a2 = *(float32*)&Arg2(row,col);
float64 e = *(float32*)&Exp(row,col);
m_b = (c == e);
_snprintf(m_acDiag, sizeof(m_acDiag)-1
, "add(Arg1,Arg2): arg1=%g, arg2=%g, Expected=%g, Actual=%g, Result: %s"
, a1, a2, e, c, BOOL_PF(m_b));
} else {
...
Pretty ugly isn't it? Using floats as args give bad output. Maybe due to using _snprintf()
?
Years ago I would use %lg
and it would be OK. Not anymore.
So, you can see here that %d is used for integers, %f for floats and %c for characters. As simple as that! %. 2f means that the variable to be printed will be of type float and '.
%s refers to a string %d refers to an integer %c refers to a character. Therefore: %s%d%s%c\n prints the string "The first character in sting ", %d prints i, %s prints " is ", and %c prints str[0].
double and float use the same format specifiers with printf
(%a
, %e
, %f
, and %g
). This is because printf
is a variadic function. Any float arguments are implicitly promoted to double before the call; you can't actually pass a float to printf
.
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