i'm试图将莱布尼兹的总和并行化,以使用pthread来逼近PI。当我运行这段代码相克的最新版本++这2个错误出现,我真的不明白为什么,我编译如下: G ++ pi2.cpp -o PI2 -lpthreadg ++错误从'void *'无效转换为'info *'[-fpermissive],错误:从'void *'无效转换为'void *(*)(void *)'[-fpermissive]
error invalid conversion from ‘void*’ to ‘info*’ [-fpermissive]
error: invalid conversion from ‘void*’ to ‘void* (*)(void*)’ [-fpermissive]
这里是我的代码:
#include <pthread.h>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#define top 1000000000
struct info
{
int inicio;
int fin;
double pi;
};
int leibniz(void *ap2){
struct info *ap;
int start, end, i;
ap = ap2;
start = ap -> inicio;
end = ap -> fin;
double x = 0.0;
double deno = 2*start + 1.0;
double nume = 1.0;
int diff = end - start;
for (i = 0; i < diff; ++i)
{
x += nume/deno;
deno += 2;
nume = -nume;
}
ap -> pi = x*4.0;
}
int main(int argc, char const *argv[])
{
int i, cont, rango;
double pi;
int numHilos = 2;
char *check = "";
if (argc >= 2)
{
numHilos = (int)strtol(argv[1], &check, 10);
if (strcmp("", check) != 0)
{
printf("Entrada invalida\n");
}
}
rango = top/numHilos;
cont = 0;
struct info array[numHilos];
for (i = 0; i < numHilos; ++i)
{
array[i].inicio = cont;
array[i].fin = rango;
cont = rango;
rango = rango + (top/numHilos);
}
pthread_t hilos[numHilos];
int r, j, *retval;
for (i = 0; i < numHilos; ++i)
{
r = pthread_create(&hilos[i], NULL, (void *)*leibniz, (void *)&array[i]);
if (r != 0)
{
perror("Error al crear el hilo");
exit(-1);
}
}
for (j = 0; j < numHilos; ++j)
{
r = pthread_join(hilos[j], (void **)&retval);
}
for (i = 0; i < numHilos; ++i)
{
pi = pi + array[i].pi;
}
printf("pi: %2.12f \n", pi);
}
在TJE开始
不像C,C++不会让你只需要给'无效*'你想要的任何东西。你必须告诉编译器你确实是这个意思,并且你没有通过强制转换来犯一个可怕的错误。 – user4581301