为什么A＆QUOT;函数名＆QUOT;评估为true在C和如何得到它警告
我最近遇到的gcc 3.2.2的以下行为写一个C程序迷迷糊糊：
I recently stumbled across the following behaviour of gcc 3.2.2 writing a c program:
在if语句我忘了函数的括号内，并写道：
In an if statement I forgot the braces of a function and wrote:
如果（myFunc的）... 而不是如果（myFunc的（））...
虽然我有pretty多一个警告打开这并没有产生错误既不是一个警告。
This did not generate an error neither a warning although I have pretty much every warning turned on.
它只是评估，以真正。
这是为什么写法律code摆在首位？
由于函数存在/有一个地址？
有谁知道如何人能避免这种错误，或者有警告选项我忽略了？在以后的版本的GCC是这个问题解决好？
It simply evaluated to true.
Why is this writing legal code in the first place ?
Because the function exists/has an address ?
Does anyone know how one could avoid such mistakes or if there is a warning option I overlooked ? Is this issue better solved in later gcc versions ?
下面确切的编译器调用的完整性：
Here the exact compiler call for completeness:
msp430-gcc -g -Os -mmcu=msp430x1611 -Wall -W -Wfloat-equal -Wundef -Wshadow -Wpointer-arith -Wbad-function-cast -Wcast-qual -Wwrite-strings -Wsign-compare -Waggregate-return -Wstrict-prototypes -Wmissing-prototypes -Wmissing-declarations
-Wredundant-decls -Wnested-externs -Wimplicit-function-declaration -Werror
（因为我被迫使用gcc 3.2.3没有-Wextra）
(Since I'm forced to use gcc 3.2.3 there is no -Wextra)
如果（myFunc的）等同于如果（安培; myFunc的），所以你测试功能，这当然将永远不为零，即地址正确的。
if (myFunc) is equivalent to if (&myFunc), so you're testing the address of a function, which of course will always be non-zero, i.e. true.
使用GCC 4.2.1和 -Wall 我得到以下警告：
With gcc 4.2.1 and -Wall I get the following warning:
myfunc.c：11：警告：myFunc的地址将始终评估为真