Use the operator feature to replace the if judgment statement

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

1. & judgment features

#include <stdio.h>int sumf(int i){    int sum = 0;#if 0    i && (sum = i+ sumf(i-1));#else    if(i!=0){        sum = i+ sumf(i-1);    }#endif    return sum;    }int main(){    printf("sum(%d)=%d\n",4,sumf(4));    printf("aaaaaaaaaaaa\n");}

 

Compiled with GCC, it seems that if statements are more efficient.

sumf:// if(i!=0){ sum = i+ sumf(i-1);    }
    pushl    %ebp    movl    %esp, %ebp    subl    $40, %esp    movl    $0, -12(%ebp)    cmpl    $0, 8(%ebp)    je    .L2    movl    8(%ebp), %eax    subl    $1, %eax    movl    %eax, (%esp)    call    sumf    addl    8(%ebp), %eax    movl    %eax, -12(%ebp).L2:    movl    -12(%ebp), %eax    leave    ret

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

    pushl    %ebp    movl    %esp, %ebp    subl    $40, %esp    movl    $0, -12(%ebp)    cmpl    $0, 8(%ebp)    je    .L3    movl    8(%ebp), %eax    subl    $1, %eax    movl    %eax, (%esp)    call    sumf    addl    8(%ebp), %eax    movl    %eax, -12(%ebp)    cmpl $0, -12(%ebp).L3:    movl    -12(%ebp), %eax    leave    ret

 

 

2. Alternative Iteration Methods :::::(! Is more practical)

# Include <stdio. h> typedef unsigned int (* Fun) (unsigned INT); unsigned int solution3_teminator (unsigned int N) {return 0;} unsigned int sum_solution3 (unsigned int N) {static Fun f [2] = {solution3_teminator, sum_solution3}; return N + F [!! N] (N-1); // note here} int sumf (int I) {int sum = 0; # If 0 I & (sum = I + sumf (I-1); # else if (I! = 0) {sum = I + sumf (I-1);} # endif return sum;} int main () {printf ("sum_solution3 (% d) = % d \ n ", 4, sum_solution3 (4); printf ("sum (% d) = % d \ n", 4, sumf (4 )); printf ("aaaaaaaaaaaa \ n ");}

 

Use the operator feature to replace the if judgment statement