Question

I find this strange.

While it makes sense that strtod accepts ‘e’ as one of the characters (exactly one to be precise) in the input string I find that it also accepts ‘d’.

Can someone please explain?

#include < stdio.h >
#include < stdlib.h >
int main ()
{
char *s[] = {"1a1", "1e1", "1d1", "1f1"};
char * pEnd;
double d0, d1, d2, d3;
d0 = strtod (s[0],&pEnd);
d1 = strtod (s[1],NULL);
d2 = strtod (s[2],NULL);
d3 = strtod (s[3],NULL);
printf ("::: [%f] [%f] [%f] [%f] \n", d0, d1, d2, d3);
return 0;
}

Answer 1

What Compiler/Libraries are you using to compile this code? Assuming you’re on Visual Studio, this behaviour is expected (quoting text from the MSDN):

strtod expects nptr to point to a
string of the following form:

[whitespace] [sign] [digits] [.digits]
[ {d | D | e | E}[sign]digits]

You can find the full documentation for strtod here

Other implementations of the library would probably support something similar. However, the man-page for strtod found here, doesn’t state that d is recognized as a valid character for the conversion purpose. In such a case, it would cause the parsing of the input string to stop and only characters parsed until that point would be converted (the same happens for the strings containing a and f).

Perhaps you should look at documentation for your library implementation and find out the format that strtod would be able to parse for your specific case.