我正在编写一个简单的RPC程序,它可以对两个int
数字进行加,减,乘,除。我写了int
然后我使用int
来生成文件。当我执行服务器,并且客户端给它两个简单的整数时,服务器给出了大的浮点数。
有什么问题?
file.x
如下:
struct entier{
int a;
int b;};
typedef struct entier params;
struct resultat{
float result;
int resp;};
typedef struct resultat res;
program CALC{
version VERS1{
void CALCUL_NULL(void)=0;
res ADDITION(params )=1;
res SOUTRACTION(params)=2;
res MULTIPLICATION(params)=3;
res DIVISION(params)=4;
}=1;
}=0x20000001;
服务器代码是:
/*
* This is sample code generated by rpcgen.
* These are only templates and you can use them
* as a guideline for developing your own functions.
*/
#include "file.h"
void *
calcul_null_1_svc(void *argp, struct svc_req *rqstp)
{
static char * result;
/*
* insert server code here
*/
return (void *) &result;
}
res *
addition_1_svc(params *argp, struct svc_req *rqstp)
{
static res result;
result.result= argp->a + argp->b;
result.resp=0;
return &result;
}
res *
soutraction_1_svc(params *argp, struct svc_req *rqstp)
{
static res result;
result.result= argp->a - argp->b;
result.resp=0;
return &result;
}
res *
multiplication_1_svc(params *argp, struct svc_req *rqstp)
{
static res result;
result.result= argp->a * argp->b;
result.resp=0;
return &result;
}
res *
division_1_svc(params *argp, struct svc_req *rqstp)
{
static res result;
if(argp->b ==0)
result.resp=1;
else
result.result= argp->a / argp->b;
return &result;
}
客户:
/*
* This is sample code generated by rpcgen.
* These are only templates and you can use them
* as a guideline for developing your own functions.
*/
#include "file.h"
int x;
int y;
void
calc_1(char *host)
{
CLIENT *clnt;
void *result_1;
char *calcul_null_1_arg;
params arguments;
arguments.a=(float)x;
arguments.b=(float)y;
res *respons;
#ifndef DEBUG
clnt = clnt_create (host, CALC, VERS1, "udp");
if (clnt == NULL) {
clnt_pcreateerror (host);
exit (1);
}
#endif /* DEBUG */
result_1 = calcul_null_1((void*)&calcul_null_1_arg, clnt);
if (result_1 == (void *) NULL) {
clnt_perror (clnt, "call failed");
}
respons = addition_1(&arguments, clnt);
printf("the additon is%f \n",respons->result);
if (respons == (res *) NULL) {
clnt_perror (clnt, "call failed");
}
respons = soutraction_1(&arguments, clnt);
printf("the substraction is %f \n",respons->result);
if (respons == (res *) NULL) {
clnt_perror (clnt, "call failed");
}
respons = multiplication_1(&arguments, clnt);
printf("the multiplication is %f \n",respons->result);
if (respons == (res *) NULL) {
clnt_perror (clnt, "call failed");
}
respons = division_1(&arguments, clnt);
if (respons == (res *) NULL) {
clnt_perror (clnt, "call failed");
if(respons->resp==1)
printf("error division by zero");
else printf("the division is %f \n",respons->result);
}
#ifndef DEBUG
clnt_destroy (clnt);
#endif /* DEBUG */
}
int
main (int argc, char *argv[])
{
char *host;
if (argc < 4) {
printf ("usage: %s server_host\n", argv[0]);
exit (1);
}
host = argv[1];
x=(int) argv[2];
y= (int) argv[3];
calc_1 (host);
exit (0);
}
执行:
blitzkrieg@blitzkrieg-TravelMate-P253:~/Desktop/rpc$ ./client localhost 5 6
the additon is2131577472.000000
the substraction is -2.000000
the multiplication is -88320960.000000
blitzkrieg@blitzkrieg-TravelMate-P253:~/Desktop/rpc$
答案 0 :(得分:0)
您似乎将entier.a
和entier.b
定义为整数,但是将它们作为浮点数传递。我希望这会在编译时生成错误,但如果没有,那么它可能会尝试对浮点数进行整数运算,这将无法正常工作。
struct entier{
int a;
int b;
};
typedef struct entier params;
和
params arguments;
arguments.a=(float)x;
arguments.b=(float)y;
问题2
x=(int) argv[2];
y= (int) argv[3];
在C中,这将采用char *
并尝试将其作为int
进行处理,这也不会达到您想要的效果。您需要使用某种转换函数将这些转换为整数。 atoi()
可以在这里做你需要的。
由于从int
到float
和char[]
到{int
的大量转换问题,您确定在编译时没有收到有关此问题的警告,因为此代码不应按原样运行{1}}。