The information in this article applies to:
SYMPTOMSThe printf() in the sample code below does not print out the correct values; the float is printed with a hexadecimal specifier before the float specifier. CAUSEThe problem is that a 2-byte format specifier is being used for an 8-byte value. RESOLUTIONTo correct this problem, replace "%x" and "%lx" with "%lx %lx". MORE INFORMATIONSample Code
The above code produces incorrect output. However, the source code is
incorrect. When printing, the above code is using a 2-byte hex format
specifier %x or a 4-byte hex format specifier %lx with an 8-byte
double argument (the float is also passed as a double). This produces
the incorrect results. To solve this problem, use two %lx format
specifiers to remove 8-bytes off the stack before printing the second
double. For example:
This works correctly for Microsoft C under MS-DOS or OS/2; however,
this code may not be portable under other systems that support types
of different sizes.
Eight bytes are passed, independent of the fact that one argument is a float and the other a double, because the float is being promoted to a double. This is because all floats are passed as doubles unless they are specified as floats in the prototype. Because printf() has a variable number of parameters, the arguments are not prototyped and therefore all floats are promoted up to doubles. Additional query words:
Keywords : kbCRT kbVC kbVC100 kbVC150 kbVC151 kbVC152 kbVC200 kbVC210 kbVC400 kbVC500 kbVC600 |
Last Reviewed: July 1, 1999 © 2000 Microsoft Corporation. All rights reserved. Terms of Use. |