C语言 将输入的数据存储在数组中
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Store data in array from input
提问by ranaarjun
I'm beginner in C. Please dont mind if my question is lame. In this program which i have written, when I use 'for' loop first time , I expect only 3 values is stored in an array but it stores 4 values and in next 'for' loop as expected show 3 values. My Question is why in 1st 'for' loop it takes 4 values instead of 3?
我是 C 初学者。请不要介意我的问题是否蹩脚。在我编写的这个程序中，当我第一次使用 'for' 循环时，我希望只有 3 个值存储在一个数组中，但它存储了 4 个值，并且在下一个 'for' 循环中按预期显示了 3 个值。我的问题是为什么在第一个“for”循环中它需要 4 个值而不是 3 个？
#include<stdio.h>
void main()
{
int marks[3];
int i;
for(i=0;i<3;i++)
{
printf("Enter a no\n");
scanf("%d\n",(marks+i));
}
for(i=0;i<3;i++)
{
printf("%d\n",*(marks+i));
}
}
回答by Sanyam Goel
\nin scanfwas the problem
\n在scanf是问题
#include<stdio.h>
int main()
{
int marks[3];
int i;
for(i=0;i<3;i++)
{
printf("Enter a no\n");
scanf("%d",(marks+i));
}
printf("\nEntered values:\n");
for(i=0;i<3;i++)
{
printf("%d\n",*(marks+i));
}
return 0;
}
Reason:
原因：
I expect only
3values is stored in an array but it stores 4 values and in next 'for' loop as expected show 3 values. My Question is why in 1st 'for' loop it takes 4 values instead of 3?
我希望只有
3值存储在数组中，但它存储 4 个值，并在下一个“for”循环中按预期显示 3 个值。我的问题是为什么在第一个“for”循环中它需要 4 个值而不是 3 个？
First:No, it only stores 3number but not 4numbers in array marks[].
第一：不，它只存储3数字而不是4数组中的数字marks[]。
Second:interesting to understand loop runs only for three times for i = 0to i < 3. The for loop runs according to condition. More interesting code is stuck in scanf()as described below:
第二：有趣的理解循环只运行三次i = 0到i < 3。for 循环根据条件运行。更有趣的代码scanf()如下所述：
Your confusion is why you have to enter four numbers, its not because you loop runs 4times but its because scanf()function returns only when you enter a non-space char (and after some enterpress you inputs a number symbol that is non-space char).
您的困惑是为什么您必须输入四个数字，这不是因为您循环运行4次数，而是因为scanf()函数仅在您输入非空格字符时才返回（并且在enter按下一些键后您输入了一个非空格字符的数字符号）。
To understand this behavior read manual: int scanf(const char *format, ...);:
要了解这种行为，请阅读手册int scanf(const char *format, ...);：
A sequence of white-space characters (space, tab, newline, etc.; see
isspace(3)). This directive matches any amount of white space, including none, in the input.
一系列空白字符（空格、制表符、换行符等；请参阅
isspace(3)）。该指令 匹配 input 中任意数量的空格，包括无空格。
Because in first for loop's, in scanf()you have included \nin format string, so scanf()returns only if press a number enter(or a non-space key).
因为在第一个 for 循环中， scanf()您已包含\n在格式字符串中，因此scanf()仅在按下数字enter（或非空格key）时才返回。
scanf("%d\n",(marks+i));
^
|
new line char
What happens?
发生什么了？
Suppose input to program is:
假设程序的输入是：
23 <--- because of %d 23 stored in marks[0] as i = 0
<enter> <--- scanf consumes \n, still in first loop
543 <--- scanf returns, and leave 542 unread,
then in next iteration 543 read by scanf in next iteration
<enter>
193
<enter> <--- scanf consumes \n, still in 3rd loop
<enter> <--- scanf consumes \n, still in 3rd loop
123 <--- remain unread in input stream
回答by Provessor
remove \nand ican be created in the if statement as for (int i = 0; i < 3; i++) {}
删除\n并i可以在 if 语句中创建为for (int i = 0; i < 3; i++) {}
