我要尝试做的事情： 我正在尝试做一个程序，以分析并提供有关任何给定数字集的一些相关信息。
为了使它不仅对我的研究有用，我正在编写它来生成任意给定数量的元素的随机数组。之后，我计划放置该选项以处理带有一组数字的文件。
我的问题是什么 我遇到的问题是，当我被要求重复，修改数组大小或退出程序时，它仅在第二个输入上做出相应的响应，而我不知道这种行为的原因是什么。
代码如下：
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>
/*This program creates an array, calculates its standard deviation and identify its large element.*/
int DV(int n){ /*DV -> Compute the standard deviation, identify the large element and its respective position*/
int i, l, p = 0;
float x[n], MVx = 0, j = 0, k = 0;
char rst;
printf("Be a random array with %d elements, created from 0 to %d: \n", n, n-1);
for(i = 0; i < n; i++){ //Generates an array with n random elements
x[i] = rand() % n;
MVx += x[i];
printf("%.0f ", x[i]);
}
MVx /= n; // Here MV assumes the Arithmetic mean
printf("\nThe arithmetic mean is: %.5f\n", MVx);
for (i = 0; i < n; i++) {
j = x[i] - MVx;
j *= j;
k += j;
}
printf("The standard deviation is %.5f\n\n", sqrt(k/n));
j = 0; // j needs to be reseted
for(i = 0, l = i + 1; i < n - 1, l < n; i++, l++){ //Compares all elements and puts its largest value on j
switch (x[i] > x[l]){
case 1: if(j < x[i]){ j = x[i]; p = i;} break;
case 0: if(j < x[l]){ j = x[l]; p = l;} break;
}
}
printf("The largest value on sample is: %.2f\nIt occurs for the first time on the %dº element\n", j, p+1);
printf("\n\nRepeat? (y/n)\nA/a to change array size: ");
while((getchar()) != '\n');
rst = getchar();
switch (rst) {
case 'N':
case 'n': return 1; break;
case 'A':
case 'a': return 2; printf("\n\n"); break;
case 'Y':
case 'y': return 3; printf("\n\n"); break;
default: return -100;
}
}
int main(void){
int n;
char rst = 'a', qtd = 'a';
while(rst == 'a') {
if(qtd == 'a'){
printf("Insert the amount of elements to be computed: ");
scanf("%d", &n);
}
srand(time(NULL)); //Generates randomic seed
DV(n);
switch(DV(n)){
case 3: printf("\n\n"); rst = 'a'; qtd = 'n'; break;
case 2: printf("\n\n"); rst = qtd = 'a'; break;
case 1: printf("\n\n"); return 0; break;
default: printf("Invalid entry\n\n"); return 0;
}
}
}
答案 0 :(得分：0)
我不知道这种情况是否还会继续发生像DV()这样的两次@Dmitri通话。但这当然是我可以对@Neil Edelman所说的内容做出的另一种解释。
当我取出DV()函数的接口解释并将提示和决策代码放在main()函数上时，程序就像假装一样工作。
我将不得不评论使用Welford的在线算法...由于它在处理大量元素时可能会表现出奇怪的行为，因此我会保留我的代码，直到对一定数量的元素都可以正常工作为止在2,090,000.00和2,100,000.00之间。而且我没有使用双重...
现在可以使用的代码如下：
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>
/*This program creates an array, calculates its standard deviation and identify its large element.*/
int DV(int n){ /*DV -> Compute the standard deviation, identify the large element and its respective position*/
int i, l, p = 0;
float x[n], MVx = 0, j = 0, k = 0;
char rst;
printf("Be a random array with %d elements, created from 0 to %d: \n", n, n-1);
for(i = 0; i < n; i++){ //Generates an array with n random elements
x[i] = rand() % n;
MVx += x[i];
printf("%.0f ", x[i]);
}
MVx /= n; // Here MV assumes the Arithmetic mean
printf("\nThe Arithmetic Mean is: %.5f\n", MVx);
for (i = 0; i < n; i++) {
j = x[i] - MVx;
j *= j;
k += j;
}
printf("The standard deviation is: %.5f\n\n", sqrt(k/n));
j = 0; // j needs to be reseted
for(i = 0, l = i + 1; i < n - 1, l < n; i++, l++){ //Compares all elements and puts its value on j
switch (x[i] > x[l]){
case 1: if(j < x[i]){ j = x[i]; p = i;} break;
case 0: if(j < x[l]){ j = x[l]; p = l;} break;
}
}
printf("The largest value on sample is: %.2f\nIt occurs for the first time on the %dº element\n", j, p+1);
}
int main(void){
int n;
char rst = 'a', qtd = 'a';
while(rst == 'a') {
if(qtd == 'a'){
printf("Insert the amount of elements to be computed: ");
scanf("%d", &n);
}
srand(time(NULL)); //Generates randomic seed
DV(n);
printf("\n\nRepeat? (y/n)\nA/a to change array size: ");
while((getchar()) != '\n');
rst = getchar();
switch (rst) {
case 'N':
case 'n': return 0; break;
case 'A':
case 'a': printf("\n\n"); rst = qtd = 'a'; break;
case 'Y':
case 'y': printf("\n\n"); rst = 'a'; qtd = 'n'; break;
default: printf("Invalid Entry\n\n"); return 0;
}
}
}
答案 1 :(得分：0)
使用C99的新功能float x[n]将数组存储在堆栈上时，请注意n较大时的堆栈溢出。参见How to declare and use huge arrays of 1 billion integers in C?和https://wiki.sei.cmu.edu/confluence/display/c/MEM05-C.+Avoid+large+stack+allocations。
如果RAND_MAX为2,147,483,647，则x = rand()返回[准]均匀分布E[x] = 1,073,741,823。假设IEEE 754是32位float，则只能准确存储16,777,217个整数，请参见Which is the first integer that an IEEE 754 float is incapable of representing exactly?。当添加n个副本时，精度下降为1/n。当n大时，这也很重要。在物理实验室中，我们总是被提醒，使用double来操纵测量几乎总是合适的，这就是为什么它很重要的原因（对于时间而言，但通常也是）https://randomascii.wordpress.com/2012/02/13/dont-store-that-in-a-float/。
此代码是Welford在线算法的实现，https://en.wikipedia.org/wiki/Algorithms_for_calculating_variance#Welford%27s_Online_algorithm（用C语言除外）。这样做的优点是您不需要知道元素的数量，也不必存储元素，（处于在线状态，内存为O(1)）；而且，从数值上讲，它比增加可能很大的和更稳定。
#include <stdlib.h> /* EXIT_ size_t */
#include <stdio.h> /* printf */
#include <math.h> /* sqrt */
/** Measurement. C version of Python
\url{ https://en.wikipedia.org/wiki/Algorithms_for_calculating_variance#Welford's_Online_algorithm }. */
struct Mx {
size_t count;
double mean, ssdm;
};
static void mx_reset(struct Mx *const measure) {
if(!measure) return;
measure->count = 0;
measure->mean = 0;
measure->ssdm = 0;
}
static void mx_add(struct Mx *const measure, const double replica) {
size_t n;
double delta;
if(!measure) return;
n = ++measure->count;
delta = replica - measure->mean;
measure->mean += delta / n;
measure->ssdm += delta * (replica - measure->mean);
}
static double mx_mean(const struct Mx *const measure) {
if(!measure || !measure->count) return NAN;
return measure->mean;
}
static double mx_sample_variance(const struct Mx *const measure) {
if(!measure || measure->count <= 1) return NAN;
return measure->ssdm / (measure->count - 1);
}
static double mx_population_variance(const struct Mx *const measure) {
if(!measure || !measure->count) return NAN;
return measure->ssdm / measure->count;
}
/** This is the example from
\url{ https://en.wikipedia.org/wiki/Standard_deviation }. */
int main(void) {
const float fulmars_f[] = { 727.7f, 1086.5f, 1091.0f, 1361.3f, 1490.5f,
1956.1f }, fulmars_m[] = { 525.8f, 605.7f, 843.3f, 1195.5f, 1945.6f,
2135.6f, 2308.7f, 2950.0f };
const size_t fulmars_f_size = sizeof fulmars_f / sizeof *fulmars_f,
fulmars_m_size = sizeof fulmars_m / sizeof *fulmars_m;
struct Mx f, m;
size_t i;
mx_reset(&f), mx_reset(&m);
/* Converts float -> double. */
for(i = 0; i < fulmars_f_size; i++) mx_add(&f, fulmars_f[i]);
for(i = 0; i < fulmars_m_size; i++) mx_add(&m, fulmars_m[i]);
printf("female breeding Northern fulmars\nmean:\t%f.\nstddev:\t%f\n"
"population stddev: %f\n\nmale breeding Northern fulmars\n"
"mean:\t%f.\nstddev:\t%f\npopulation stddev: %f\n", mx_mean(&f),
sqrt(mx_sample_variance(&f)), sqrt(mx_population_variance(&f)),
mx_mean(&m), sqrt(mx_sample_variance(&m)),
sqrt(mx_population_variance(&m)));
return EXIT_SUCCESS;
}
在这种情况下，用户界面是一个常数，但是对某些估算的for(i = 0; i < n; i++) mx_add(rand() % n);进行n会将用户界面与统计计算分开。