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我有一个文件reken.x,client.c和server.c来创建一个分布式系统。客户端将两个素数(如此一个数字)的乘积发送到服务器上的“ontbind”函数,该函数将数字分解为两个素数。当我在“reken_in”和“reken_uit”结构中声明变量“getal1”,“antwoord1”和“antwoord2”作为整数时,这可以正常工作。然而,当我使用unsigned long类型多头出现以下错误:使用无符号长整型与rpcgen给出错误
[email protected]:~/Desktop/tarbal$ rpcgen reken.x
unsigned long long getal1;
^^^^^^^^^^^^^^^^^^^
reken.x, line 2: expected '*' or 'identifier'
我如何与无符号long long类型这项工作?请参阅下面的我的代码。
reken.x
struct reken_in { /*input argument*/
unsigned long long getal1;
};
struct reken_uit {
unsigned long long antwoord1;
unsigned long long antwoord2;
};
program BEREKEN {
version BERVERS{
reken_uit ONTBIND(reken_in) =1; /*procedure nr. 1*/
}=1;
}=0x31230000; /*programma nummer*/
client.c
#include <stdio.h>
#include <rpc/rpc.h>
#include "reken.h"
#include <stdlib.h>
main(int argc,char *argv[])
{
//hi
CLIENT *cl;
char *server;
reken_in getallen;
reken_uit *antw;
if(argc !=3) {
puts("aantal argumenten niet goed <server productPriemen>");
exit(1);
}
server=argv[1];
getallen.getal1= strtoull(argv[2], NULL, 10);
cl=clnt_create(server,BEREKEN,BERVERS,"tcp");
if(cl==NULL) {
printf("fout bij het zoeken naar de server");
clnt_pcreateerror(server);
exit(1);
}
antw= ontbind_1(&getallen,cl);
if(antw==NULL)
puts("fout bij teruggeef parameter");
printf("Het getal %llu is het product van priemgetal %llu en %llu\n",getallen.getal1, antw->antwoord1, antw->antwoord2);
exit (0);
}
server.c
#include <stdio.h>
#include <rpc/rpc.h>
#include <dirent.h>
#include "reken.h"
#include <math.h>
reken_uit *ontbind_1_svc(struct reken_in *g,struct svc_req *reg)
{
static reken_uit getallen;
unsigned long long number = g->getal1;
unsigned long long i;
unsigned long long fact;
for(i = 2; i <= sqrt(number); i++) { //loop tot de sqrt
if (!isPrime(i) || (i % 2 == 0 && i != 2)) continue;
if (number % i != 0) continue;
fact = number/i; //deel door de i. De i is een prime
int prime = isPrime(fact);
if (!prime) continue;
getallen.antwoord1 = i;
getallen.antwoord2 = fact;
return (&getallen);
}
return (&getallen);
}
int isPrime (const long long number) {
int i;
for(i = 2; i < number/2; i++) { // 7
if ((number % i) == 0) // 7 % 2 == 1
return 0;
}
return 1;
}