[已解决]请老师指点va_star()第二个参数是函数的最后一个参数名.而不是参数的数量哦

wow7jiao · 发表于 2018-7-2 21:26:27

本帖最后由 wow7jiao 于 2018-7-2 23:41 编辑

http://bbs.fishc.com/thread-77359-1-1.html

#include <stdio.h>
#include <stdarg.h>

int sum(int n, ...);

int sum(int n, ...) // 三个小点是占位符，表示参数个数不确定
{
      int i, sum = 0;
      va_list vap; // 定义参数列表

      va_start(vap, n); // 初始化参数列表，如果是 int sum(int gg, ...); 则这里应该是 va_start(vap, gg);<-------------我看来看去这里举例gg就是参数个数，没有明白那上面“函数的最后一个参数名”是什么意思？
      for (i = 0; i < n; i++)
      {
            sum += va_arg(vap, int); // 获取参数值
      }
      va_end(vap); // 首尾工作，关闭参数列表

      return sum;
}

int main()
{
      int result;

      result = sum(3, 1, 2, 3);

      printf("result = %d\n", result);

      return 0;
}

最佳答案

月排行榜 / 总排行榜

人造人

2018-7-2 21:26:28

请站在汇编语言的角度看问题

#include <stdio.h>
#include <stdarg.h>
int sum0(int n, ...)
{
int sum = 0;
void *p;
p = &n;
for(int i = 0; i < n; ++i)
{
++(int *)p;
sum += *(int *)p;
}
p = NULL;
return sum;
}
int sum1(int n, ...)
{
int sum = 0;
va_list vap;
va_start(vap, n);
for(int i = 0; i < n; ++i)
{
sum += va_arg(vap, int);
}
va_end(vap);
return sum;
}
int main(void)
{
printf("%d\n", sum0(5, 100, 200, 300, 400, 500));
printf("%d\n", sum1(5, 100, 200, 300, 400, 500));
return 0;
}

复制代码

来看看这两个函数有多相似

int sum0(int n, ...)
{
0137F290 55 push ebp
0137F291 8B EC mov ebp,esp
0137F293 81 EC E4 00 00 00 sub esp,0E4h
0137F299 53 push ebx
0137F29A 56 push esi
0137F29B 57 push edi
0137F29C 8D BD 1C FF FF FF lea edi,[ebp-0E4h]
0137F2A2 B9 39 00 00 00 mov ecx,39h
0137F2A7 B8 CC CC CC CC mov eax,0CCCCCCCCh
0137F2AC F3 AB rep stos dword ptr es:[edi]
int sum = 0;
0137F2AE C7 45 F8 00 00 00 00 mov dword ptr [sum],0
void *p;
p = &n;
0137F2B5 8D 45 08 lea eax,[n]
0137F2B8 89 45 EC mov dword ptr [p],eax
for(int i = 0; i < n; ++i)
0137F2BB C7 45 E0 00 00 00 00 mov dword ptr [ebp-20h],0
0137F2C2 EB 09 jmp sum0+3Dh (0137F2CDh)
0137F2C4 8B 45 E0 mov eax,dword ptr [ebp-20h]
0137F2C7 83 C0 01 add eax,1
0137F2CA 89 45 E0 mov dword ptr [ebp-20h],eax
0137F2CD 8B 45 E0 mov eax,dword ptr [ebp-20h]
0137F2D0 3B 45 08 cmp eax,dword ptr [n]
0137F2D3 7D 16 jge sum0+5Bh (0137F2EBh)
{
++(int *)p;
0137F2D5 8B 45 EC mov eax,dword ptr [p]
0137F2D8 83 C0 04 add eax,4
0137F2DB 89 45 EC mov dword ptr [p],eax
sum += *(int *)p;
0137F2DE 8B 45 EC mov eax,dword ptr [p]
0137F2E1 8B 4D F8 mov ecx,dword ptr [sum]
0137F2E4 03 08 add ecx,dword ptr [eax]
0137F2E6 89 4D F8 mov dword ptr [sum],ecx
}
0137F2E9 EB D9 jmp sum0+34h (0137F2C4h)
p = NULL;
0137F2EB C7 45 EC 00 00 00 00 mov dword ptr [p],0
return sum;
0137F2F2 8B 45 F8 mov eax,dword ptr [sum]
}
0137F2F5 5F pop edi
0137F2F6 5E pop esi
0137F2F7 5B pop ebx
0137F2F8 8B E5 mov esp,ebp
0137F2FA 5D pop ebp
0137F2FB C3 ret

复制代码

int sum1(int n, ...)
{
00FAF8F0 55 push ebp
00FAF8F1 8B EC mov ebp,esp
00FAF8F3 81 EC E4 00 00 00 sub esp,0E4h
00FAF8F9 53 push ebx
00FAF8FA 56 push esi
00FAF8FB 57 push edi
00FAF8FC 8D BD 1C FF FF FF lea edi,[ebp-0E4h]
00FAF902 B9 39 00 00 00 mov ecx,39h
00FAF907 B8 CC CC CC CC mov eax,0CCCCCCCCh
00FAF90C F3 AB rep stos dword ptr es:[edi]
int sum = 0;
00FAF90E C7 45 F8 00 00 00 00 mov dword ptr [sum],0
va_list vap;
va_start(vap, n);
00FAF915 8D 45 0C lea eax,[ebp+0Ch]
00FAF918 89 45 EC mov dword ptr [vap],eax
for(int i = 0; i < n; ++i)
00FAF91B C7 45 E0 00 00 00 00 mov dword ptr [ebp-20h],0
00FAF922 EB 09 jmp sum1+3Dh (0FAF92Dh)
00FAF924 8B 45 E0 mov eax,dword ptr [ebp-20h]
00FAF927 83 C0 01 add eax,1
00FAF92A 89 45 E0 mov dword ptr [ebp-20h],eax
00FAF92D 8B 45 E0 mov eax,dword ptr [ebp-20h]
00FAF930 3B 45 08 cmp eax,dword ptr [n]
00FAF933 7D 17 jge sum1+5Ch (0FAF94Ch)
{
sum += va_arg(vap, int);
00FAF935 8B 45 EC mov eax,dword ptr [vap]
00FAF938 83 C0 04 add eax,4
00FAF93B 89 45 EC mov dword ptr [vap],eax
00FAF93E 8B 4D EC mov ecx,dword ptr [vap]
00FAF941 8B 55 F8 mov edx,dword ptr [sum]
00FAF944 03 51 FC add edx,dword ptr [ecx-4]
00FAF947 89 55 F8 mov dword ptr [sum],edx
}
00FAF94A EB D8 jmp sum1+34h (0FAF924h)
va_end(vap);
00FAF94C C7 45 EC 00 00 00 00 mov dword ptr [vap],0
return sum;
00FAF953 8B 45 F8 mov eax,dword ptr [sum]
}
00FAF956 5F pop edi
00FAF957 5E pop esi
00FAF958 5B pop ebx
00FAF959 8B E5 mov esp,ebp
00FAF95B 5D pop ebp
00FAF95C C3 ret

复制代码

跳转到最佳答案楼层

人造人 · 发表于 2018-7-2 21:26:28

请站在汇编语言的角度看问题

#include <stdio.h>
#include <stdarg.h>
int sum0(int n, ...)
{
int sum = 0;
void *p;
p = &n;
for(int i = 0; i < n; ++i)
{
++(int *)p;
sum += *(int *)p;
}
p = NULL;
return sum;
}
int sum1(int n, ...)
{
int sum = 0;
va_list vap;
va_start(vap, n);
for(int i = 0; i < n; ++i)
{
sum += va_arg(vap, int);
}
va_end(vap);
return sum;
}
int main(void)
{
printf("%d\n", sum0(5, 100, 200, 300, 400, 500));
printf("%d\n", sum1(5, 100, 200, 300, 400, 500));
return 0;
}

复制代码

来看看这两个函数有多相似

int sum0(int n, ...)
{
0137F290 55 push ebp
0137F291 8B EC mov ebp,esp
0137F293 81 EC E4 00 00 00 sub esp,0E4h
0137F299 53 push ebx
0137F29A 56 push esi
0137F29B 57 push edi
0137F29C 8D BD 1C FF FF FF lea edi,[ebp-0E4h]
0137F2A2 B9 39 00 00 00 mov ecx,39h
0137F2A7 B8 CC CC CC CC mov eax,0CCCCCCCCh
0137F2AC F3 AB rep stos dword ptr es:[edi]
int sum = 0;
0137F2AE C7 45 F8 00 00 00 00 mov dword ptr [sum],0
void *p;
p = &n;
0137F2B5 8D 45 08 lea eax,[n]
0137F2B8 89 45 EC mov dword ptr [p],eax
for(int i = 0; i < n; ++i)
0137F2BB C7 45 E0 00 00 00 00 mov dword ptr [ebp-20h],0
0137F2C2 EB 09 jmp sum0+3Dh (0137F2CDh)
0137F2C4 8B 45 E0 mov eax,dword ptr [ebp-20h]
0137F2C7 83 C0 01 add eax,1
0137F2CA 89 45 E0 mov dword ptr [ebp-20h],eax
0137F2CD 8B 45 E0 mov eax,dword ptr [ebp-20h]
0137F2D0 3B 45 08 cmp eax,dword ptr [n]
0137F2D3 7D 16 jge sum0+5Bh (0137F2EBh)
{
++(int *)p;
0137F2D5 8B 45 EC mov eax,dword ptr [p]
0137F2D8 83 C0 04 add eax,4
0137F2DB 89 45 EC mov dword ptr [p],eax
sum += *(int *)p;
0137F2DE 8B 45 EC mov eax,dword ptr [p]
0137F2E1 8B 4D F8 mov ecx,dword ptr [sum]
0137F2E4 03 08 add ecx,dword ptr [eax]
0137F2E6 89 4D F8 mov dword ptr [sum],ecx
}
0137F2E9 EB D9 jmp sum0+34h (0137F2C4h)
p = NULL;
0137F2EB C7 45 EC 00 00 00 00 mov dword ptr [p],0
return sum;
0137F2F2 8B 45 F8 mov eax,dword ptr [sum]
}
0137F2F5 5F pop edi
0137F2F6 5E pop esi
0137F2F7 5B pop ebx
0137F2F8 8B E5 mov esp,ebp
0137F2FA 5D pop ebp
0137F2FB C3 ret

复制代码

int sum1(int n, ...)
{
00FAF8F0 55 push ebp
00FAF8F1 8B EC mov ebp,esp
00FAF8F3 81 EC E4 00 00 00 sub esp,0E4h
00FAF8F9 53 push ebx
00FAF8FA 56 push esi
00FAF8FB 57 push edi
00FAF8FC 8D BD 1C FF FF FF lea edi,[ebp-0E4h]
00FAF902 B9 39 00 00 00 mov ecx,39h
00FAF907 B8 CC CC CC CC mov eax,0CCCCCCCCh
00FAF90C F3 AB rep stos dword ptr es:[edi]
int sum = 0;
00FAF90E C7 45 F8 00 00 00 00 mov dword ptr [sum],0
va_list vap;
va_start(vap, n);
00FAF915 8D 45 0C lea eax,[ebp+0Ch]
00FAF918 89 45 EC mov dword ptr [vap],eax
for(int i = 0; i < n; ++i)
00FAF91B C7 45 E0 00 00 00 00 mov dword ptr [ebp-20h],0
00FAF922 EB 09 jmp sum1+3Dh (0FAF92Dh)
00FAF924 8B 45 E0 mov eax,dword ptr [ebp-20h]
00FAF927 83 C0 01 add eax,1
00FAF92A 89 45 E0 mov dword ptr [ebp-20h],eax
00FAF92D 8B 45 E0 mov eax,dword ptr [ebp-20h]
00FAF930 3B 45 08 cmp eax,dword ptr [n]
00FAF933 7D 17 jge sum1+5Ch (0FAF94Ch)
{
sum += va_arg(vap, int);
00FAF935 8B 45 EC mov eax,dword ptr [vap]
00FAF938 83 C0 04 add eax,4
00FAF93B 89 45 EC mov dword ptr [vap],eax
00FAF93E 8B 4D EC mov ecx,dword ptr [vap]
00FAF941 8B 55 F8 mov edx,dword ptr [sum]
00FAF944 03 51 FC add edx,dword ptr [ecx-4]
00FAF947 89 55 F8 mov dword ptr [sum],edx
}
00FAF94A EB D8 jmp sum1+34h (0FAF924h)
va_end(vap);
00FAF94C C7 45 EC 00 00 00 00 mov dword ptr [vap],0
return sum;
00FAF953 8B 45 F8 mov eax,dword ptr [sum]
}
00FAF956 5F pop edi
00FAF957 5E pop esi
00FAF958 5B pop ebx
00FAF959 8B E5 mov esp,ebp
00FAF95B 5D pop ebp
00FAF95C C3 ret

复制代码

BngThea · 发表于 2018-7-3 08:43:40

应该是指的是一个变量，如果直接表示个数的话，常量就行了，或者该函数设计的时候就没必要加入该形参

wow7jiao · 发表于 2018-7-3 10:54:38

https://blog.csdn.net/al86866365/article/details/56485178

好像意思 int n 是从栈里弹出的最后一个参数，作为结束地址

wow7jiao · 发表于 2018-7-3 13:48:17

人造人发表于 2018-7-3 11:14
请站在汇编语言的角度看问题

谢谢，人造人老师，我明白了，小甲鱼是说va_start()第二个参数要填变量n（上面写什么下面写什么），而不能填常量，例如5。&n是取一个标记地址。

PS:人造人老师，第12行代码 ++（int *）p，我用dev翻译显示错误不是左值，晚上我再回去vs试试。

人造人 · 发表于 2018-7-3 18:01:34

wow7jiao 发表于 2018-7-3 13:48
谢谢，人造人老师，我明白了，小甲鱼是说va_start()第二个参数要填变量n（上面写什么下面写什么），而不 ...

现在呢

#include <stdio.h>
#include <stdarg.h>
int sum0(int n, ...)
{
int sum = 0;
void *p;
p = &n;
for(int i = 0; i < n; ++i)
{
//++(int *)p;
p = (int *)p + 1;
sum += *(int *)p;
}
p = NULL;
return sum;
}
int sum1(int n, ...)
{
int sum = 0;
va_list vap;
va_start(vap, n);
for(int i = 0; i < n; ++i)
{
sum += va_arg(vap, int);
}
va_end(vap);
return sum;
}
int main(void)
{
printf("%d\n", sum0(5, 100, 200, 300, 400, 500));
printf("%d\n", sum1(5, 100, 200, 300, 400, 500));
return 0;
}

复制代码

wow7jiao · 发表于 2018-7-3 21:46:07

人造人发表于 2018-7-3 18:01
现在呢

++（int *）p这里p为什么是不是左值

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