c返回int值不起作用

pol*_*nux -1 c function return-value

这是我的代码:

#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <pthread.h>

int num_mezzo_1(int num_orig);
int num_mezzo_2(int num_orig);

int main(int argc, char *argv[]){
    int num,contatore,tmp=0,tmp_1,tmp_2;
    num=atoi(argv[1]);
    if(num <= 3){
        printf("%d è primo\n", num);
        exit(0);
    }
    else{
        num_mezzo_1(num);
        num_mezzo_2(num);
        tmp=tmp_1+tmp_2;
            //using printf to debug
        printf("t1 %d, t2 %d\n", tmp_1,tmp_2);
        if(tmp>2){
            printf("%d NON è primo\n", num);
        }
        else{
            printf("%d è primo\n", num);
        }
    }
    exit(0);
}

int num_mezzo_1(int num_orig){
    int tmp_1=0,cont_1;
    for(cont_1=1; cont_1<=(num_orig/2); cont_1++){
        if((num_orig % cont_1) == 0){
            tmp_1++;
        }
    }
    //using printf to debug
    printf("\n%d\n", tmp_1);
    return tmp_1;
}

int num_mezzo_2(int num_orig){
    int tmp_2=0,cont_2;
    for(cont_2=((num_orig/2)+1); cont_2<=num_orig; cont_2++){
        if((num_orig % cont_2) == 0){
            tmp_2++;
        }
    }
    //using printf to debug
    printf("\n%d\n\n", tmp_2);
    return tmp_2;
}
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该程序计算数字是否为素数.
如果我给数字13作为输入,该函数num_1具有值1tmp_1和功能num_2具有价值1tmp_2和二者都是正确的.
问题是,tmp=tmp_1+tmp_2返回一个很大的巨大价值,我不明白为什么.

mat*_*975 6

您所呼叫的函数num_mezzo_1()num_mezzo_2(),但你不保存其返回值,让你的变量tmp_1,并tmp_2保持未初始化.

编辑:尝试更改代码

    num_mezzo_1(num);
    num_mezzo_2(num);
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    tmp_1 = num_mezzo_1(num);
    tmp_2 = num_mezzo_2(num);
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在else块中,看看你是否得到了你所期望的.

工作代码:

tmp=(num_mezzo_1(num)+num_mezzo_2(num));
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