void call(int x,int y,int z)
{
printf("%d %d %d",x,y,z);
}
int main()
{
int a=10;
call(a,a++,++a);
return 0;
}
该方案对不同编汇人提供不同产出,在编集时,总会因任何原因,把产出编成一栏。
void call(int x,int y,int z)
{
printf("%d %d %d",x,y,z);
}
int main()
{
int a=10;
call(a,a++,++a);
return 0;
}
该方案对不同编汇人提供不同产出,在编集时,总会因任何原因,把产出编成一栏。
由于行为不明确。 编辑可在任何顺序上评价<代码>a,a++
和++a
。 (从技术上讲,由于我们援引未经界定的行为,它实际上没有做任何事情;它可能撰写它所希望的任何法典。) 根据他们重新评价的命令,结果有所不同。
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