blocks|key|656386|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|656387|scanf("%25[%5E\n]s",name);|code-block|syntax|javascript|656388|\n只是为扫描的字符串设置分隔符。|offset|length|style|CODE|656389|entityMap^0|0|0|0|2|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Try this:

<pre><code>scanf("%[^\n]s",name);
</code></pre>

<code>\n</code> just sets the delimiter for the scanned string.

blocks|key|656415|text|下面是一个示例，说明如何使用fgets函数获取包含空格的输入。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|656416|#include+<stdio.h>

int+main()
{
++++char+name[100];
++++printf("Enter+your+name:+");
++++fgets(name,+100,+stdin);+
++++printf("Your+Name+is:+%25s",+name);
++++return+0;
}|code-block|syntax|javascript|656417|entityMap^0|E|5|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Here is an example of how you can get input containing spaces by using the <code>fgets</code> function. 

<pre><code>#include &lt;stdio.h&gt;

int main()
{
 char name[100];
 printf("Enter your name: ");
 fgets(name, 100, stdin); 
 printf("Your Name is: %s", name);
 return 0;
}
</code></pre>

blocks|key|873801|text|scanf("+%25[%5E\t\n]s",&str);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|873802|str是从中获取字符串的变量。|unstyled|offset|length|style|CODE|873803|entityMap^0|0|0|3|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>scanf(" %[^\t\n]s",&amp;str);
</code></pre>

<code>str</code> is the variable in which you are getting the string from.

blocks|key|656751|text|注意:使用fgets()时，当您在CLI+(命令行解释器)中使用fgets()作为小输入时，数组中的最后一个字符将是'\n‘，因为您以'Enter’结束字符串。因此，当您打印字符串时，编译器在打印字符串时总是转到下一行。如果您希望输入字符串具有类似于null终止的字符串行为，请使用这个简单的技巧。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|656752|#include<stdio.h>
int+main()
{
+int+i,size;
+char+a[100];
+fgets(a,100,stdin);;
+size+=+strlen(a);
+a[size-1]='\0';

return+0;
}|code-block|syntax|javascript|656753|更新:在其他用户的帮助下更新。|656754|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

NOTE: When using fgets(), the last character in the array will be '\n' at times when you use fgets() for small inputs in CLI (command line interpreter) , as you end the string with 'Enter'. So when you print the string the compiler will always go to the next line when printing the string. If you want the input string to have null terminated string like behavior, use this simple hack.

<pre><code>#include&lt;stdio.h&gt;
int main()
{
 int i,size;
 char a[100];
 fgets(a,100,stdin);;
 size = strlen(a);
 a[size-1]='\0';

return 0;
}
</code></pre>

Update: Updated with help from other users.

blocks|key|874152|text|正确的答案是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|874153|#include+<stdio.h>

int+main(void)
{
++++char+name[100];

++++printf("Enter+your+name:+");
++++//+pay+attention+to+the+space+in+front+of+the+%25
++++//that+do+all+the+trick
++++scanf("+%25[%5E\n]s",+name);
++++printf("Your+Name+is:+%25s",+name);

++++return+0;
}|code-block|syntax|javascript|874154|%25前面空格非常重要，因为如果您的程序中有另外几个scanf，假设您有一个整数值的scanf和另一个具有双精度值的scanf+...当您到达char+(字符串名)的scanf时，该命令将被跳过，并且您不能为其输入值...但是如果你把那个空格放在%25前面，一切都是可以的，不会跳过任何东西。|874155|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

The correct answer is this:

<pre><code>#include &lt;stdio.h&gt;

int main(void)
{
 char name[100];

 printf("Enter your name: ");
 // pay attention to the space in front of the %
 //that do all the trick
 scanf(" %[^\n]s", name);
 printf("Your Name is: %s", name);

 return 0;
}
</code></pre>

That space in front of % is very important, because if you have in your program another few scanf let's say you have 1 scanf of an integer value and another scanf with a double value... when you reach the scanf for your char (string name) that command will be skipped and you can't enter value for it... but if you put that space in front of % will be ok everything and not skip nothing.

blocks|key|874171|text|#include+<stdio.h>
//+read+a+line+into+str,+return+length
int+read_line(char+str[])+{
int+c,+i=0;
c+=+getchar();
while+(c+!=+'\n'+&&+c+!=+EOF)+{+
+++str[i]+=+c;
+++c+=+getchar();
+++i%2B%2B;
}
str[i]+=+'\0';
return+i;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|874172|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>#include &lt;stdio.h&gt;
// read a line into str, return length
int read_line(char str[]) {
int c, i=0;
c = getchar();
while (c != '\n' &amp;&amp; c != EOF) { 
 str[i] = c;
 c = getchar();
 i++;
}
str[i] = '\0';
return i;
}
</code></pre>

blocks|key|874106|text|使用此代码，您可以接受输入，直到按下键盘的enter键为止。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|874107|char+ch[100];
int+i;
for+(i+=+0;+ch[i]+!=+'\n';+i%2B%2B)
{
++++scanf("%25c+",+&ch[i]);
}|code-block|syntax|javascript|874108|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Using this code you can take input till pressing enter of your keyboard. 

<pre><code>char ch[100];
int i;
for (i = 0; ch[i] != '\n'; i++)
{
 scanf("%c ", &amp;ch[i]);
}
</code></pre>

blocks|key|874212|text|虽然上面提到的方法确实有效，但每种方法都有自己的问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|874213|如果您使用的是支持posix的平台，则可以使用getline()或getdelim()。如果你使用windows和minigw作为你的编译器，那么它应该是可用的。|offset|length|style|CODE|874214|getline()定义为：|874215|ssize_t+getline(char+**lineptr,+size_t+*n,+FILE+*stream);|874216|为了接受输入，首先你需要创建一个指向char类型的指针。|874217|#include+<stdio.h>
#include<stdlib.h>

//+s+is+a+pointer+to+char+type.
char+*s;
//+size+is+of+size_t+type,+this+number+varies+based+on+your+guess+of+
//+how+long+the+input+is,+even+if+the+number+is+small,+it+isn't+going+
//+to+be+a+problem
size_t+size+=+10;

int+main(){
//+allocate+s+with+the+necessary+memory+needed,+%2B1+is+added+
//+as+its+input+also+contains,+/n+character+at+the+end.
++++s+=+(char+*)malloc(size%2B1);
++++getline(&s,&size,stdin);
++++printf("%25s",s);
++++return+0;
}|code-block|syntax|javascript|874218|示例输入：Hello+world+to+the+world!|874219|输出：Hello+world+to+the+world!\n|874220|这里需要注意的一件事是，即使为s分配的内存是11字节，而当输入大小是26字节时，getline也会使用realloc()重新分配s。|874221|所以你的输入有多长并不重要。|874222|使用读取的no.of字节更新size，根据上面的示例，输入size将为27。|874223|getline()还将\n视为input.So，您的's‘将在末尾保留'\n’。|874224|还有一个更通用的getline()版本，也就是getdelim()，它有一个额外的参数，即delimiter。|874225|getdelim()定义为：|874226|ssize_t+getdelim(char+**lineptr,+size_t+*n,+int+delim,+FILE+*stream);|874227|Linux+man+page|874228|entityMap|0|LINK|mutability|MUTABLE|url|http://man7.org/linux/man-pages/man3/getline.3.html^0|0|N|9|X|A|0|0|9|0|0|1L|0|0|0|5|P|0|3|R|0|1F|9|1S|1|0|0|E|4|T|4|Z|2|0|0|9|B|2|0|8|9|N|A|19|9|0|0|A|0|0|1X|0|0|E|0|0^^$0|@$1|2|3|4|5|6|7|1K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1L|8|@$D|1M|E|1N|F|G]|$D|1O|E|1P|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|1Q|8|@$D|1R|E|1S|F|G]]|9|@]|A|$]]|$1|J|3|K|5|6|7|1T|8|@$D|1U|E|1V|F|G]]|9|@]|A|$]]|$1|L|3|M|5|6|7|1W|8|@]|9|@]|A|$]]|$1|N|3|O|5|P|7|1X|8|@]|9|@]|A|$Q|R]]|$1|S|3|T|5|6|7|1Y|8|@$D|1Z|E|20|F|G]]|9|@]|A|$]]|$1|U|3|V|5|6|7|21|8|@$D|22|E|23|F|G]]|9|@]|A|$]]|$1|W|3|X|5|6|7|24|8|@$D|25|E|26|F|G]|$D|27|E|28|F|G]]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|29|8|@]|9|@]|A|$]]|$1|10|3|11|5|6|7|2A|8|@$D|2B|E|2C|F|G]|$D|2D|E|2E|F|G]|$D|2F|E|2G|F|G]]|9|@]|A|$]]|$1|12|3|13|5|6|7|2H|8|@$D|2I|E|2J|F|G]|$D|2K|E|2L|F|G]]|9|@]|A|$]]|$1|14|3|15|5|6|7|2M|8|@$D|2N|E|2O|F|G]|$D|2P|E|2Q|F|G]|$D|2R|E|2S|F|G]]|9|@]|A|$]]|$1|16|3|17|5|6|7|2T|8|@$D|2U|E|2V|F|G]]|9|@]|A|$]]|$1|18|3|19|5|6|7|2W|8|@$D|2X|E|2Y|F|G]]|9|@]|A|$]]|$1|1A|3|1B|5|6|7|2Z|8|@]|9|@$D|30|E|31|1|32]]|A|$]]|$1|1C|3|-4|5|6|7|33|8|@]|9|@]|A|$]]]|1D|$1E|$5|1F|1G|1H|A|$1I|1J]]]]

While the above mentioned methods do work, but each one has it's own kind of problems.

You can use <code>getline()</code> or <code>getdelim()</code>, if you are using posix supported platform.
If you are using windows and minigw as your compiler, then it should be available.

<code>getline()</code> is defined as :

<code>ssize_t getline(char **lineptr, size_t *n, FILE *stream);</code>

In order to take input, first you need to create a pointer to char type.

<pre><code>#include &lt;stdio.h&gt;
#include&lt;stdlib.h&gt;

// s is a pointer to char type.
char *s;
// size is of size_t type, this number varies based on your guess of 
// how long the input is, even if the number is small, it isn't going 
// to be a problem
size_t size = 10;

int main(){
// allocate s with the necessary memory needed, +1 is added 
// as its input also contains, /n character at the end.
 s = (char *)malloc(size+1);
 getline(&amp;s,&amp;size,stdin);
 printf("%s",s);
 return 0;
}
</code></pre>

Sample Input:<code>Hello world to the world!</code>

Output:<code>Hello world to the world!\n</code>

One thing to notice here is, even though allocated memory for s is 11 bytes,
where as input size is 26 bytes, getline reallocates <code>s</code> using <code>realloc()</code>.

So it doesn't matter how long your input is.

<code>size</code> is updated with no.of bytes read, as per above sample input <code>size</code> will be <code>27</code>.

<code>getline()</code> also considers <code>\n</code> as input.So your 's' will hold '\n' at the end.

There is also more generic version of <code>getline()</code>, which is <code>getdelim()</code>, which takes one more extra argument, that is <code>delimiter</code>.

<code>getdelim()</code> is defined as:

<code>ssize_t getdelim(char **lineptr, size_t *n, int delim, FILE *stream);</code>

<a href="http://man7.org/linux/man-pages/man3/getline.3.html" rel="nofollow noreferrer">Linux man page</a>

blocks|key|874276|text|如果需要读取多行数据，则需要清除buffer。示例：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|874277|int+n;
scanf("%25d",+&n);
char+str[1001];
char+temp;
scanf("%25c",&temp);+//+temp+statement+to+clear+buffer
scanf("%25[%5E\n]",str);|code-block|syntax|javascript|874278|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you need to read more than one line, need to clear buffer. Example:

<pre><code>int n;
scanf("%d", &amp;n);
char str[1001];
char temp;
scanf("%c",&amp;temp); // temp statement to clear buffer
scanf("%[^\n]",str);
</code></pre>

blocks|key|873640|text|"%25s"将读取输入，直到到达空格。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|873641|如果您想读取一行(即到达换行符之前的所有字符，包括空格)，gets可能是一个很好的起点。|873642|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cplusplus.com/reference/clibrary/cstdio/gets/^0|0|4|0|T|4|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@]|D|@$9|T|A|U|1|V]]|E|$]]|$1|H|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

<code>"%s"</code> will read the input until whitespace is reached.

<a href="http://www.cplusplus.com/reference/clibrary/cstdio/gets/" rel="nofollow">gets</a> might be a good place to start if you want to read a line (i.e. all characters including whitespace until a newline character is reached).

blocks|key|656638|text|“巴拉克·奥巴马”在“巴拉克”和“奥巴马”之间有一个空格。为了适应这一点，使用下面的代码；|type|unstyled|depth|inlineStyleRanges|entityRanges|data|656639|#include+<stdio.h>
int+main()
{
++++printf("Enter+your+name\n");
+++char+a[80];
+++gets(a);
+++printf("Your+name+is+%25s\n",+a);
+++return+0;
}|code-block|syntax|javascript|656640|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

"Barack Obama" has a space between 'Barack' and 'Obama'. To accommodate that, use this code;

<pre><code>#include &lt;stdio.h&gt;
int main()
{
 printf("Enter your name\n");
 char a[80];
 gets(a);
 printf("Your name is %s\n", a);
 return 0;
}
</code></pre>

blocks|key|873884|text|scanf("%25s",name);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|873885|对scanf输入使用&|unstyled|offset|length|style|CODE|873886|entityMap^0|0|1|5|A|1|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]|$G|Q|H|R|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|S|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>scanf("%s",name);
</code></pre>

use <code>&amp;</code> with <code>scanf</code> input

I'm using Ubuntu and I'm also using Geany and CodeBlock as my IDE.
What I'm trying to do is reading a string (like <code>"Barack Obama"</code>) and put it in a variable:

<pre class="lang-c prettyprint-override"><code>#include &lt;stdio.h&gt;

int main(void)
{
 char name[100];

 printf("Enter your name: ");
 scanf("%s", name);
 printf("Your Name is: %s", name);

 return 0;
}
</code></pre>

Output:

<pre class="lang-none prettyprint-override"><code>Enter your name: Barack Obama
Your Name is: Barack
</code></pre>

How can I make the program read the whole name?

Reading string from input with space character?

集成开发环境

编译器

解释器

Ubuntu

我使用Ubuntu，我也使用Geany和CodeBlock作为我的集成开发环境。我尝试做的是读取一个字符串(如"Barack Obama")并将其放入一个变量中：#include <stdio.h>int main(void){ char name[100]; printf("Enter your name: "); scanf("%s", name); printf("Y

问是否使用空格字符从输入中读取字符串？
EN

回答 12

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问是否使用空格字符从输入中读取字符串？EN

回答 12

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问是否使用空格字符从输入中读取字符串？
EN