Consider this little program: #include <stdio.h> int main() { char c = 0xFF; printf(\"%d\\n\", c); return 0;
In pure ansi C, is there any way to show that, given char c1 = 1, c2 = 2; the type of the following: c1 + c2
Hello all I hope you can help me understand why getopt used an int and the handling of the optopt variable in getopt. Pretty new to C++.
Until rec开发者_如何学编程ently, I\'d considered the decision by most systems implementors/vendors to keep plain int 32-bit even on 64-bit machines a sort of expedient wart. With modern C99 fixed-size
I was having开发者_如何学编程 a look over this page: http://www.devbistro.com/tech-interview-questions/Cplusplus.jsp, and didn\'t understand this question:
#include<stdio.h> int main(void) { char c = 0x80; printf(\"%d\\n\", c << 1); return 0; } The output is -256 in this case. If I write c << 0 then the outp开发者_如何转开发ut is -128
When comparing a type larger than int, with an integer constant, should I place the constant on the left or the right to ensure the correct comparison is performed?
Normally, C requires that a binary operator\'s operands are promoted to the type of the higher-ranking operand. This can be exploited to avoid filling code with verbose casts, for example:
I learned from ----As to when default promotions kick in: default argument promotions are used exactly when the expected type of the argument is unknown, which is to say when ther开发者_高级运维e\'s n
I need to copy files in one directory to another directory where the lastwritetime is greater than or equal to 3/26/2010 9:00pm. 开发者_如何学Go I\'m using:
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