我有一个函数,我想将其作为参数传递给另一个函数(我们称之为funX)。这里的funX原型:将函数指针传递给函数+函数参数
void funX(const unsigned char *, unsigned char *, size_t, const somestruct *, unsigned char *, const int);
和我的功能(可以称之为讨人喜欢),其中要求funX:
unsigned char * funY(unsigned char *in, unsigned char *out, size_t len, unsigned char *i, void *k, int ed, void (*f)(unsigned char *, unsigned char *, size_t, const void *, unsigned char *, const int))
{
f(in, out, len, k, i, ed);
}
但我有一些警告在编译:
test.c: In function ‘main’:
test.c:70:5: warning: passing argument 7 of ‘funY’ from incompatible pointer type [enabled by default]
test.c:11:17: note: expected ‘void (*)(unsigned char *, unsigned char *, size_t, const void *, unsigned char *, const int)’ but argument is of type ‘void (*)(const unsigned char *, unsigned char *, size_t, const struct somestruct *, unsigned char *, const int)’
的签名使用的typedef的建议[这里] (http://stackoverflow.com/a/9143434/ 841108)。这使得更容易地声明函数参数。 –
错误消息看起来非常清晰,函数指针类型不对应于要传递的函数。 –
“funY”声明中的函数指针参数与“funX”的类型不同。只是第一个区别:第一个参数是'funX',是一个'const unsigned char *',而函数指针的第一个参数是'unsigned char *'。 –