blocks|key|772969|text|代码中存在多个问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|772970|772971|major+char+string1[3]+=+"abc";定义了大小为3字节的目标数组，该数组没有足够的空间来容纳字符串"abc"，包括其空终止符，而且绝对没有足够长的时间在最后从string2接收额外的字符。将此更改为char+string1[7]+=+"abc";|unordered-list-item|offset|length|style|BOLD|CODE|772972|major+char+string2[3]+=+"def";定义了大小为3字节的源数组，该数组没有足够的空间用于字符串"def"+(包括其空终止符)，因此将不会为空"def"+string1[strlen(string1)+-+1]+=+'\0';覆盖abc的最后一个字符，与necessary.|772973|major不同的是，string2[strlen(string2)+-+1]+=+'\0';|772974|minor+string2的相同注释应该定义为void+strconcatenate(char+*string1,+char+*string2)|772975|minor+i中的const+char+*，而j应该定义为size_t|772976|minor类型，以便与标准函数strcat兼容，strconcatenate应该有返回类型char+*并返回指向目标数组的指针。但是，正如Jonathan所评论的，返回值的一个更有用的设计是在连接字符串的末尾返回一个指向null的指针。您已经知道字符串的起始位置；这不是非常有趣的信息。但是知道连接后字符串的结尾在哪里--这是有用的信息，只有连接函数才知情。-|772977|772978|以下是修改后的版本：|772979|#include+<stdio.h>
#include+<string.h>

char+*strconcatenate(char+*string1,+const+char+*string2)+{
++++char+*p+=+string1+%2B+strlen(string1);
++++while+((*p+=+*string1)+!=+'\0')+{
++++++++string1%2B%2B;
++++++++p%2B%2B;
++++}
++++return+string1;++/*+strcat+semantics+*/
++++//return+p;++++++/*+alternative+semantics+*/
}

int+main()+{
++++char+string1[7]+=+"abc";
++++char+string2[]+=+"def";

++++strconcatenate(string1,+string2);
++++printf("Resultant+string+=+%25s\n",+string1);
++++return+0;
}|code-block|syntax|javascript|772980|entityMap^0|0|0|0|5|6|O|1P|5|2J|7|33|O|0|0|5|6|O|1N|5|2A|5|2G|10|3I|3|0|0|5|A|10|0|0|5|6|7|N|1D|0|0|5|6|1|9|C|N|1|T|6|0|0|5|F|6|O|E|19|6|0|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|15|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|16|8|@$F|17|G|18|H|I]|$F|19|G|1A|H|J]|$F|1B|G|1C|H|J]|$F|1D|G|1E|H|J]|$F|1F|G|1G|H|J]]|9|@]|A|$]]|$1|K|3|L|5|E|7|1H|8|@$F|1I|G|1J|H|I]|$F|1K|G|1L|H|J]|$F|1M|G|1N|H|J]|$F|1O|G|1P|H|J]|$F|1Q|G|1R|H|J]|$F|1S|G|1T|H|J]]|9|@]|A|$]]|$1|M|3|N|5|E|7|1U|8|@$F|1V|G|1W|H|I]|$F|1X|G|1Y|H|J]]|9|@]|A|$]]|$1|O|3|P|5|E|7|1Z|8|@$F|20|G|21|H|I]|$F|22|G|23|H|J]|$F|24|G|25|H|J]]|9|@]|A|$]]|$1|Q|3|R|5|E|7|26|8|@$F|27|G|28|H|I]|$F|29|G|2A|H|J]|$F|2B|G|2C|H|J]|$F|2D|G|2E|H|J]|$F|2F|G|2G|H|J]]|9|@]|A|$]]|$1|S|3|T|5|E|7|2H|8|@$F|2I|G|2J|H|I]|$F|2K|G|2L|H|J]|$F|2M|G|2N|H|J]|$F|2O|G|2P|H|J]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|2Q|8|@]|9|@]|A|$]]|$1|V|3|W|5|6|7|2R|8|@]|9|@]|A|$]]|$1|X|3|Y|5|Z|7|2S|8|@]|9|@]|A|$10|11]]|$1|12|3|-4|5|6|7|2T|8|@]|9|@]|A|$]]]|13|$]]

There are multiple problems in your code:
<ul>
<li>[major] <code>char string1[3] = &quot;abc&quot;;</code> defines the destination array with a size of 3 bytes which does not have a enough space for the string <code>&quot;abc&quot;</code> including its null terminator and definitely not long enough to receive the extra characters from <code>string2</code> at the end. Change this to <code>char string1[7] = &quot;abc&quot;;</code></li>
<li>[major] <code>char string2[3] = &quot;def&quot;;</code> defines the source array with a size of 3 bytes which does not have a enough space for the string <code>&quot;def&quot;</code> including its null terminator, hence will not be null terminated.</li>
<li>[major] <code>string1[strlen(string1) - 1] = '\0';</code> overwrites the last character of <code>abc</code>, this not necessary.</li>
<li>[major] same remark for <code>string2[strlen(string2) - 1] = '\0';</code></li>
<li>[minor] <code>string2</code> should be defined as <code>const char *</code> in <code>void strconcatenate(char *string1, char *string2)</code></li>
<li>[minor] <code>i</code> and <code>j</code> should be defined with type <code>size_t</code></li>
<li>[minor] for compatibility with the standard function <code>strcat</code>, <code>strconcatenate</code> should have a return type of <code>char *</code> and return the pointer to the destination array. Yet, As commented by Jonathan Leffler, a more useful design for the return value is to return a pointer to the null at the end of the concatenated string. You already know where the start of the string is; that isn't very interesting information. But knowing where the end of the string is after concatenation — that is useful information which only the concatenate function is privy to. –</li>
</ul>
Here is a modified version:
<pre><code>#include &lt;stdio.h&gt;
#include &lt;string.h&gt;

char *strconcatenate(char *string1, const char *string2) {
 char *p = string1 + strlen(string1);
 while ((*p = *string1) != '\0') {
 string1++;
 p++;
 }
 return string1; /* strcat semantics */
 //return p; /* alternative semantics */
}

int main() {
 char string1[7] = &quot;abc&quot;;
 char string2[] = &quot;def&quot;;

 strconcatenate(string1, string2);
 printf(&quot;Resultant string = %s\n&quot;, string1);
 return 0;
}
</code></pre>

blocks|key|1318894|text|因为string1和string2有3个字符，所以您需要用4个字节(+'\0‘)来声明它，当您不需要从strlen中减去1时，因为它指向’\0‘字符。下面是对代码的更改。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1318895|int+main()
{
++char+string1[4]+=+"abc";
++string1[strlen(string1)]+=+'\0';
++char+string2[4]+=+"def";
++string2[strlen(string2)]+=+'\0';

++strconcatenate(string1,string2);++
++printf("Resultant+string+=+%25s\n",string1);
++return+0;
}|code-block|syntax|javascript|1318896|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|$]]

Since string1 and string2 has 3 characters so you need to declare it with 4 bytes('\0') and also when you need not to subtract 1 from strlen because it points to '\0' character.
Below is the change to the code.
<pre><code>int main()
{
 char string1[4] = &quot;abc&quot;;
 string1[strlen(string1)] = '\0';
 char string2[4] = &quot;def&quot;;
 string2[strlen(string2)] = '\0';

 strconcatenate(string1,string2); 
 printf(&quot;Resultant string = %s\n&quot;,string1);
 return 0;
}
</code></pre>

blocks|key|1631368|text|在main中，您用三个元素声明了字符数组。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1631369|char+string1[3]+=+"abc";
//...
char+string2[3]+=+"def";|code-block|syntax|javascript|1631370|这些数组不包含字符串，因为它们不存储用作初始化器的字符串文本的终止零字符'\0'。第一个数组包含三个字符{+'a',+'b',+'c'+}，第二个数组包含这三个字符{+'d',+'e',+'f'+}。|offset|length|style|CODE|1631371|然后，将标准字符串函数strlen应用于这些数组，期望传递给它的数组包含字符串。也就是说，函数通过计数字符来计算字符串中的字符数，直到遇到终止零字符为止。由于数组不包含终止的零字符'\0‘，也就是说数组不包含字符串，所以strlen调用调用未定义的行为。|1631372|如果要将一个字符串附加到另一个字符串，则目标字符数组的大小应足以容纳第二个字符串。|1631373|因此，函数main可以如下所示|1631374|int+main(+void+)+{
++++char+string1[7]+=+"abc";
++++char+string2[]+=+"def";

++++printf(+"Resultant+string+=+%25s\n",+strconcatenate(+string1,+string2+)+);

++++return+0;
}|1631375|现在，两个数组string1和string2都包含字符串。此外，数组string1保留足够的空间来容纳存储在数组string2中的字符串。|1631376|函数strconcatenate的声明应该类似于标准C函数strcat的声明。|1631377|char+*+strconcatenate(+char+*string1,+const+char+*string2+);|1631378|也就是说，第二个参数应该有限定符const，因为传递的数组在函数中没有改变，函数应该返回包含连接字符串的目标数组。|1631379|函数定义如下所示|1631380|char+*+strconcatenate(+char+*string1,+const+char+*string2+)+
{
++++for+(+char+*p+=+string1+%2B+strlen(+string1+);+*p%2B%2B+=+*string2%2B%2B;+);

++++return+string1;
}|1631381|entityMap^0|0|0|10|4|1G|H|2A|H|0|B|6|32|6|0|0|0|0|7|7|F|7|Y|7|1K|7|0|2|E|T|6|0|0|G|5|0|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1A|8|@$I|1B|J|1C|K|L]|$I|1D|J|1E|K|L]|$I|1F|J|1G|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|1H|8|@$I|1I|J|1J|K|L]|$I|1K|J|1L|K|L]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1M|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1N|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|1O|8|@]|9|@]|A|$E|F]]|$1|U|3|V|5|6|7|1P|8|@$I|1Q|J|1R|K|L]|$I|1S|J|1T|K|L]|$I|1U|J|1V|K|L]|$I|1W|J|1X|K|L]]|9|@]|A|$]]|$1|W|3|X|5|6|7|1Y|8|@$I|1Z|J|20|K|L]|$I|21|J|22|K|L]]|9|@]|A|$]]|$1|Y|3|Z|5|D|7|23|8|@]|9|@]|A|$E|F]]|$1|10|3|11|5|6|7|24|8|@$I|25|J|26|K|L]]|9|@]|A|$]]|$1|12|3|13|5|6|7|27|8|@]|9|@]|A|$]]|$1|14|3|15|5|D|7|28|8|@]|9|@]|A|$E|F]]|$1|16|3|-4|5|6|7|29|8|@]|9|@]|A|$]]]|17|$]]

In main you declared character arrays with three elements
<pre><code>char string1[3] = &quot;abc&quot;;
//...
char string2[3] = &quot;def&quot;;
</code></pre>
These arrays do not contain strings because they do not store the terminating zero character <code>'\0'</code> of the string literals used as initializers. The first array contains three characters <code>{ 'a', 'b', 'c' }</code> and the second array contains these three characters <code>{ 'd', 'e', 'f' }</code>.
Then you applied the standard string function <code>strlen</code> to these arrays which expects that passed to it arrays contains strings. That is the function calculates the number of characters in a string by counting characters until the terminating zero character is encountered. As the arrays do not contain the terminating zero character '\0' that is as the arrays do not contain strings the calls of <code>strlen</code> invoke undefined behavior.
If you want to append one string to another string then the destination character array shall have enough size to be able to accommodate the second string.
Thus the function main can look the following way
<pre><code>int main( void ) {
 char string1[7] = &quot;abc&quot;;
 char string2[] = &quot;def&quot;;

 printf( &quot;Resultant string = %s\n&quot;, strconcatenate( string1, string2 ) );

 return 0;
}
</code></pre>
Now the both arrays <code>string1</code> and <code>string2</code> contain strings. Moreover the array <code>string1</code> reserved enough space to accommodate the string stored in the array <code>string2</code>.
The declaration of the function <code>strconcatenate</code> should be similar to the declaration of the standard C function <code>strcat</code>.
<pre><code>char * strconcatenate( char *string1, const char *string2 );
</code></pre>
That is the second parameter should have the qualifier <code>const</code> because the passed array is not changed within the function and the function should return the destination array that will contain the concatenated strings.
The function definition will look the following way
<pre><code>char * strconcatenate( char *string1, const char *string2 ) 
{
 for ( char *p = string1 + strlen( string1 ); *p++ = *string2++; );

 return string1;
}
</code></pre>

I need to concatenate the strings <code>&quot;abc&quot;</code> and <code>&quot;def&quot;</code> using <code>strlen</code> but without using <code>strcat</code>. Can someone show me what is wrong with my <code>main()</code> function?
<pre><code>#include &lt;stdio.h&gt;
#include &lt;string.h&gt;

void strconcatenate(char *string1, char *string2) {
 int i;
 int j = strlen(string1);
 for (i = 0; string2[i]; i++) {
 string1[i+j] = string2[i];
 }
 string1[i + j]= '\0';
}

int main() {
 char string1[3] = &quot;abc&quot;;
 string1[strlen(string1) - 1] = '\0';
 char string2[3] = &quot;def&quot;;
 string2[strlen(string2) - 1] = '\0';

 strconcatenate(string1, string2); 
 printf(&quot;Resultant string = %s\n&quot;, string1);
 return 0;
}
</code></pre>

Function to concatenate two strings in C using strlen but without using strcat

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我需要使用"abc"和"def"连接字符串，使用strlen，但不使用strcat。有人能告诉我我的main()函数有什么问题吗？#include <stdio.h>#include <string.h>void strconcatenate(char *string1, char *string2) { int...

问函数使用strlen连接C中的两个字符串，但不使用strcat。
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问函数使用strlen连接C中的两个字符串，但不使用strcat。EN

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问函数使用strlen连接C中的两个字符串，但不使用strcat。
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