用于计算用户给出的n个数字的平均值的程序。
好的，所以我有这个程序的目的是你上面已经读过的。它的输出不太对劲。我弄清楚问题是什么，但找不到解决方案，因为我不是编程的新手（实际上是新手）。这是代码：
#include <stdio.h>
int main(void) {
char user_data[100];
long int sum = 0;
double average;
unsigned int numbers_count = 0;
for (int i = 0; i <= 99; ++i)
user_data[i] = 0;
unsigned int numbers[100];
for (int i = 0; i <= 99; ++i)
numbers[i] = 0;
printf("Please enter the numbers:");
fgets(user_data, sizeof(user_data), stdin);
int i = 0;
while (user_data[i] != 0) {
sscanf(user_data, "%u", &numbers[i]);
++i;
}
i = 0;
while (numbers[i] != 0) {
sum += numbers[i];
++i;
}
i = 0;
while (numbers[i] != 0) {
++numbers_count;
++i;
}
average = (float)sum / (float)numbers_count;
printf("\n\nAverage of the entered numbers is: %f",average);
return 0;
}
现在问题来了。
当我输入一个整数23时，它会以两个单独的字节存储到user_data中。我添加了一个循环来打印user_data[i]的值以找出错误。
i = 0;
while (i <= 99) {
printf("%c\n",user_data[i]);
++i;
}`
结果就是这个
这是第一个问题，这是第二个问题。 我添加了另一个与上面相同的循环来打印存储在数字[100]中的数字并找出错误，这是输出。这是一个样本
现在我的主要问题是
如何从user_data中提取完整数字？
答案 0 :(得分：2)
我认为在{23}的user_data之后布局fgets()可能会有所帮助（假设Linux或Mac新行）：
![numbers stored in numbers[]](https://i.stack.imgur.com/Ev6Ba.png){kind=link}
+-----+-----+----+----+
| '2' | '3' | \n | \0 | .....
+-----+-----+----+----+
0 1 2 3
请注意，user_data[0] 不包含2（数字2）！它包含'2'（字符'2'），其代码（再次，假设Linux）0x32（十六进制或十进制50）。
这就是为什么您尝试打印user_data[]的值并不富有成效的原因：您试图打印数字的表示，而不是数字本身。
要将该字符串转换为它所代表的整数，您可以执行以下操作：
num = atoi(user_data)
函数atoi()为您完成工作。一个更灵活的函数是strtol()，它对long int执行相同的操作（并且还可以处理表示基数中不是10的数字的字符串）。
我希望这能回答您的问题：如何从user_data中提取完整数字？
还有一些其他要点应该清理和简化代码，但是如果需要帮助，可以打开另一个问题。
答案 1 :(得分：0)
试试这个：
#include<stdio.h>
#include<stdlib.h>
int main(void)
{
int i;
char user_data[100];
long int sum = 0;
double average;
unsigned int numbers_count = 0;
for( i=0; i<= 99; ++i)
user_data[i] = 0;
unsigned int numbers[100];
for( i=0; i<= 99; ++i)
numbers[i] = 0;
printf("Please enter the numbers:");
fgets(user_data,sizeof(user_data),stdin);
//int p=0;//use with strtol(see further code)
i = 0;
int j;//this will store each number in numbers array.. so this is also the count of numbers stored - 1
for(j=0;;){
for(i=0;i<strlen(user_data);i++)
{
if(user_data[i]=='\n'){
break;
}
if(user_data[i]==' '){
j++;
i++;
//p=i;//to be used with strtol
}
numbers[j]=(numbers[j]*10)+((user_data[i]-48));//alternatively use => numbers[j]=strtol(user_data+p,NULL,10);
}
break;
}
i = 0;
while( i<=j)
{
sum += numbers[i];
++i;
}
average = (float)sum/(j+1);
printf("\n\nAverage of the entered numbers is: %f",average);
return 0;
}
示例输入
10 11 12
示例输出
11.00000000
我已经展示了两种方法来解决这个问题：
答案 2 :(得分：0)
关于发布的代码：
如果其中一个数字为零，会发生什么？
如果数字的总和超过＆＃39; sum＆＃39;
的容量，会发生什么#include <stdio.h> // sscanf(), fgets(), printf()
#include <stdlib.h> // strtol()
#include <string.h> // strtok()
// eliminate the 'magic' number by giving it a meaningful name
#define MAX_INPUTS 100
int main(void)
{
// the array can be initialized upon declaration
// which eliminates the 'for()' loop to initialize it
// char user_data[100];
// and
// initialization not actually needed as
// the call to 'fgets()' will overlay the array
// and 'fgets()' always appends a NUL byte '\0'
char user_data[ MAX_INPUTS ];
long int sum = 0;
double average;
// following variable not needed
// unsigned int numbers_count = 0;
// following code block not needed when
// 'user_data[]' initialized at declaration
// for (int i = 0; i <= 99; ++i)
// user_data[i] = 0;
// not needed, see other comments
//unsigned int numbers[100];
// not needed, as 'numbers' is eliminated
// for (int i = 0; i <= 99; ++i)
// numbers[i] = 0;
printf("Please enter the numbers:");
// should be checking the returned value
// to assure it is not NULL
// And
// this call to 'fgets()' is expecting
// all the numbers to be on a single input line
// so that could be a problem
fgets(user_data, sizeof(user_data), stdin);
// the following two code blocks will not extract the numbers
// for a number of reasons including that 'sscanf()'
// does not advance through the 'user_data[]' array
// int i = 0;
// while (user_data[i] != 0) {
// sscanf(user_data, "%u", &numbers[i]);
// ++i;
// }
// i = 0;
// while (numbers[i] != 0) {
// sum += numbers[i];
// ++i;
// }
// suggest the following,
// which also eliminates the need for 'numbers[]'
// note: the literal " \n" has both a space and a newline
// because the user is expected to enter numbers,
// separated by a space and
// 'fgets()' also inputs the newline
int i = 0;
char *token = strtok( user_data, " \n");
while( token )
{
// not everyone likes 'atoi()'
// mostly because there is no indication of any error event
// suggest using: 'strtol()'
//sum += atoi( token );
sum += strtol( token, NULL, 10 ) // could add error checking
i++;
token = strtok( NULL, " \n" );
}
// the value assigned to 'numbers_count'
// is already available in 'i'
// suggest eliminate the following code block
// and
// eliminate the 'numbers_count' variable
// i = 0;
// while (numbers[i] != 0) {
// ++numbers_count;
// ++i;
// }
// 'average' is declared as a 'double',
// so the casting should be to 'double'
// and
// if incorporating the prior comment about 'numbers_count'
// average = (float)sum / (float)numbers_count;
average = (double)sum / (double)i'
// to have the text immediately displayed on the terminal
// place a '\n' at the end of the format string.
// without adding the '\n' the text only displays
// as the program exits
// printf("\n\nAverage of the entered numbers is: %f",average);
printf("\n\nAverage of the entered numbers is: %f\n",average);
return 0;
} // end function: main