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1成tmp_1和功能num_2具有价值1入tmp_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.
您所呼叫的函数num_mezzo_1()和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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至
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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