《品悟C》疑问

zhou_jingzhi

问题1:69页
const int = 5;
int *p = (int *)&n;
++*p;
...
for(i = 0;i<n-1;i++)
{
...
}
再也没提过p，请问这个p是干什么用的？

问题2:73页，字符串比较大小，1大于2resu是正数，等于是0，小于是负数。
char s1[100],s2[100];
gets(s1);
gets(s2);
i = 0;
while((s1[i] == s2[i])&&(s1[i]!='\0'))i++;
if(s1[i == '\0'&&s2[i]=='\0'])
   resu = 0;
else
   resu = s1[i] - s2[i];

请问s1[i]和s2[i]不是代表字符么？最后一行代码，俩字符相减能算出2个字符串的大小么？

谢谢！

folklore

@pmerofc

1. 
   
2. const int n= 5;
   
3. int *p = (int *)&n;
   
4. ++*p;
   
5. ...
   
6. for(i = 0;i<n-1;i++)
   
7. {
   
8. ...
   
9. }
   
10. 
    

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is it possible the variant "n" be put in ".rdata" section (read-only memory block)?

1. 
   
2. char s1[100],s2[100];
   
3. gets(s1);
   
4. gets(s2);
   
5. i = 0;
   
6. while((s1[i] == s2[i])&&(s1[i]!='\0'))i++;
   
7. if(s1[i == '\0'&&s2[i]=='\0'])
   
8.     resu = 0;
   
9. else
   
10.     resu = s1[i] - s2[i];
    

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change to

1. 
   
2. char s1[100],s2[100];
   
3. gets(s1);
   
4. gets(s2);
   
5. i = 0;
   
6. while((s1[i] == s2[i])&&(s1[i]!='\0'))i++;
   
7. resu = s1[i] - s2[i];
   

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may be better..., be aware of the loop condition

bruceteen

1。为了简化代码
我更喜欢 ++*(int*)&n;
或者是 ++(int&)n;

2。我都不明白你哪里不明白
不过 if else 不必要把，似乎可以直接 return s1[i]-s2[i];

bruceteen

folklore 发表于 2012-10-13 14:48
@pmerofcis it possible the variant "n" be put in ".rdata" section (read-only memory block)?change to ...

这种强制转化由程序员自己负全责
就是不放在只读区，也可能被编译时优化

bruceteen

@folklore @pmerofc
我发现个问题
    const int n = 5;
    ++(int&)n;
在VC++9.0中编译失败，报：error C2105: '++' needs l-value
按道理，虽然逻辑不通，但语法没问题呀

bruceteen

回复 8# pmerofc
是C++写法，g++4.7正确，vc++9.0不行

folklore

回复 6# bruceteen


    change to

1. 
   
2.         int n = 5;
   
3.         ++(int&)n;
   
4.         printf("%d",n);
   

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and it works. i suggest the variant n be optimized to a really constant in such case.