为什么这句话会消除“未引用的函数参数警告”?

时间:2016-06-08 17:18:35

标签: c warnings compiler-warnings sdcc

“SDCC编译器用户指南”中的

我阅读了以下内容:

void to_buffer( unsigned char c )
{
c; // to avoid warning: unreferenced function argument
__asm
; save used registers here.
; If we were still using r2,r3 we would have to push them here.
; if( head != (unsigned char)(tail-1) )
mov a,_tail
dec a
xrl a,_head
; we could do an ANL a,#0x0f here to use a smaller buffer (see below)
jz t_b_end$
;
; buf[ head++ ] = c;
mov a,dpl ; dpl holds lower byte of function argument
mov dpl,_head ; buf is 0x100 byte aligned so head can be used directly
mov dph,#(_buf>>8)
movx @dptr,a
inc _head
; we could do an ANL _head,#0x0f here to use a smaller buffer (see above)
t_b_end$:
; restore used registers here
__endasm;
}

我不明白句子"c; // to avoid warning: unreferenced function argument"是什么意思,它是SDCC的特殊用法吗?还是C语言的特殊用法?

1 个答案:

答案 0 :(得分:0)

如果您有一个不使用的传入函数参数,编译器会发出警告。在您的情况下,c; void 操作,用于访问变量并避免警告。它类似于

 int func(char c)
 {
  (void)c;
  //c is never used in the function
 }

FWIW,在gcc中,启用警告,使用-Wunused-parameter选项。 (<{>在-Wextra 中启用)