利用 操作符特性 代替if判断语句

参考：http://blog.csdn.net/speedme/article/details/22916181

1.&&的判断特性

#include <stdio.h>
 
int sumf(int i)
{
    int sum = ;
#if 0
    i && (sum = i+ sumf(i-));
#else
    if(i!=){
        sum = i+ sumf(i-);
    }
#endif
    return sum;
}
 
int main()
{
    printf("sum(%d)=%d\n",,sumf());
    printf("aaaaaaaaaaaa\n");
}

用gcc编译，看起来if判断语句效率还高一点。

sumf:// if(i!=0){ sum = i+ sumf(i-1);    }
    pushl    %ebp
    movl    %esp, %ebp
    subl    $, %esp
    movl    $, -(%ebp)
    cmpl    $, (%ebp)
    je    .L2
    movl    (%ebp), %eax
    subl    $, %eax
    movl    %eax, (%esp)
    call    sumf
    addl    (%ebp), %eax
    movl    %eax, -(%ebp)
.L2:
    movl    -(%ebp), %eax
    leave
    ret

sumf://i && (sum = i+ sumf(i-1));

    pushl    %ebp
    movl    %esp, %ebp
    subl    $, %esp
    movl    $, -(%ebp)
    cmpl    $, (%ebp)
    je    .L3
    movl    (%ebp), %eax
    subl    $, %eax
    movl    %eax, (%esp)
    call    sumf
    addl    (%ebp), %eax
    movl    %eax, -(%ebp)
    cmpl    $0, -12(%ebp)
.L3:
    movl    -(%ebp), %eax
    leave
    ret

2.另类的迭代法：：：：：（!的判断特性这个比较实用一点）

#include <stdio.h>
typedef unsigned int (*fun)(unsigned int);
 
unsigned int Solution3_Teminator(unsigned int n)
{
    return ;
}
 
unsigned int Sum_Solution3(unsigned int n)
{
    static fun f[] = {Solution3_Teminator, Sum_Solution3};
    return n + f[!!n](n - ); //注意此处
}
 
int sumf(int i)
{
    int sum = ;
#if 0
    i && (sum = i+ sumf(i-));
#else
    if(i!=){
        sum = i+ sumf(i-);
    }
#endif
    return sum;
}
 
int main()
{
    printf("Sum_Solution3(%d)=%d\n",,Sum_Solution3());
    printf("sum(%d)=%d\n",,sumf());
    printf("aaaaaaaaaaaa\n");
}