Pointers 使用C语言中的引用调用查找标准差、平均值和和的程序
这就是代码:Pointers 使用C语言中的引用调用查找标准差、平均值和和的程序,pointers,standard-deviation,Pointers,Standard Deviation,这就是代码: #include <stdio.h> #include <math.h> void calculateSD(float *data,float *standardDeviation, float *sum, float *average) { int i; for(i=0; i<10; ++i) { *sum= *sum+data[i]; } *average = *sum/10; for(i=0; i<10; ++i) {
#include <stdio.h>
#include <math.h>
void calculateSD(float *data,float *standardDeviation, float *sum, float *average)
{
int i;
for(i=0; i<10; ++i)
{
*sum= *sum+data[i];
}
*average = *sum/10;
for(i=0; i<10; ++i)
{
*standardDeviation += pow(data[i] - *average, 2);
}
*standardDeviation=*standardDeviation/10;
}
int main()
{
int i;
float average=0.0;
float SD=0.0;
float sum=0.0;
float *standardDeviation=&SD;
float *psum=∑
float *paverage=&average;
float *data=(float*)calloc(5,sizeof(float));
printf("Enter 5 elements: ");
for(i=0; i < 5; ++i)
{
scanf("%f",&data[i]);
}
calculateSD(data,standardDeviation,psum,paverage);
printf("\nStandard Deviation = %f", SD);
printf("\nAverage= %f", average);
printf("\nSum= %f",sum);
return 0;
}
#包括
#包括
无效计算(浮动*数据、浮动*标准偏差、浮动*总和、浮动*平均值)
{
int i;
对于(i=0;i您的数组大小为5
float *data=(float*)calloc(5,sizeof(float));
但是当你在函数中遍历它时,你会给它10作为它的大小
for(i=0; i<10; ++i)
{
*sum= *sum+data[i];
}
for(i=0;我工作得很好!谢谢