blocks|key|816153|text|这里有一个非常简单的函数，可以将数字转换为任意基(仅限于可用数字符号的数目)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|816154|char+*reverse(char+*ptr,+size_t+length)
{
++++char+*str+=+ptr;
++++if(ptr+&&+length)
++++{
++++++++char+*end+=+ptr+%2B+length+-+1;
++++++++while(end+>+str)
++++++++{
++++++++++++char+tmp+=+*str;
++++++++++++*str%2B%2B+=+*end;
++++++++++++*end--+=+tmp;
++++++++}
++++}
++++return+ptr;
}


char+*convert(char+*buff,+long+long+val,+unsigned+base)
{
++++const+char+digits[]+=+"0123456789ABCDEFGHIJKLMNOPQRSTUVXYZ";
++++int+negative+=+val+<+0;
++++size_t+count+=+0;

++++if(base+<+2+%7C%7C+base+>+sizeof(digits)+-+1)+return+NULL;

++++do
++++{
++++++++buff[count%2B%2B]+=+digits[abs(val+%25+base)];
++++++++val+/=+base;
++++}while(val);
++++if(negative)+buff[count%2B%2B]+=+'-';
++++buff[count]+=+0;
++++return+reverse(buff,+count);
++++
}|code-block|syntax|javascript|816155|https://godbolt.org/z/Wz15h4x41|offset|length|816156|在行动部的评论之后：|816157|函数，它将根据asUnsigned参数打印值的有符号表示或无符号表示：|style|CODE|816158|char+*convert(char+*buff,+long+long+val,+unsigned+base,+int+asUnsigned)
{
++++const+char+digits[]+=+"0123456789ABCDEFGHIJKLMNOPQRSTUVXYZ";
++++int+negative+=+val+<+0;
++++size_t+count+=+0;

++++if(base+<+2+%7C%7C+base+>+sizeof(digits)+-+1)+return+NULL;

++++do
++++{
++++++++buff[count%2B%2B]+=+digits[asUnsigned+?+((unsigned)val)+%25+base+:+abs(val+%25+base)];
++++++++val+=+(asUnsigned+?+(unsigned)val+:+val)+/+base;
++++}while(val);
++++if(!asUnsigned+&&+negative)+buff[count%2B%2B]+=+'-';
++++buff[count]+=+0;
++++return+reverse(buff,+count);
++++
}|816159|https://godbolt.org/z/MdMGerTTs|816160|int+main(void)+{
++++char+str[32];
++++printf("%25s\n",+convert(str,+-15,+8,+1));++
++++printf("%25s\n",+convert(str,+-15,+8,+0));++

}|816161|结果：|816162|37777777761
-17|816163|entityMap|0|LINK|mutability|MUTABLE|url|1^0|0|0|0|V|0|0|0|7|A|0|0|0|V|1|0|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]]|$1|G|3|H|5|6|7|1A|8|@]|9|@$I|1B|J|1C|1|1D]]|A|$]]|$1|K|3|L|5|6|7|1E|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|1F|8|@$I|1G|J|1H|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|1I|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|1J|8|@]|9|@$I|1K|J|1L|1|1M]]|A|$]]|$1|U|3|V|5|D|7|1N|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|1O|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|D|7|1P|8|@]|9|@]|A|$E|F]]|$1|10|3|-4|5|6|7|1Q|8|@]|9|@]|A|$]]]|11|$12|$5|13|14|15|A|$16|H]]|17|$5|13|14|15|A|$16|T]]]]

Here is a very simple function to convert number to any base (limited to the number of available digit symbols).
<pre><code>char *reverse(char *ptr, size_t length)
{
 char *str = ptr;
 if(ptr &amp;&amp; length)
 {
 char *end = ptr + length - 1;
 while(end &gt; str)
 {
 char tmp = *str;
 *str++ = *end;
 *end-- = tmp;
 }
 }
 return ptr;
}


char *convert(char *buff, long long val, unsigned base)
{
 const char digits[] = &quot;0123456789ABCDEFGHIJKLMNOPQRSTUVXYZ&quot;;
 int negative = val &lt; 0;
 size_t count = 0;

 if(base &lt; 2 || base &gt; sizeof(digits) - 1) return NULL;

 do
 {
 buff[count++] = digits[abs(val % base)];
 val /= base;
 }while(val);
 if(negative) buff[count++] = '-';
 buff[count] = 0;
 return reverse(buff, count);
 
}
</code></pre>
<a href="https://godbolt.org/z/Wz15h4x41" rel="nofollow noreferrer">https://godbolt.org/z/Wz15h4x41</a>
After OPs comment:
function which will print signed or unsigned representation of the value depending on the <code>asUnsigned</code> parameter:
<pre><code>char *convert(char *buff, long long val, unsigned base, int asUnsigned)
{
 const char digits[] = &quot;0123456789ABCDEFGHIJKLMNOPQRSTUVXYZ&quot;;
 int negative = val &lt; 0;
 size_t count = 0;

 if(base &lt; 2 || base &gt; sizeof(digits) - 1) return NULL;

 do
 {
 buff[count++] = digits[asUnsigned ? ((unsigned)val) % base : abs(val % base)];
 val = (asUnsigned ? (unsigned)val : val) / base;
 }while(val);
 if(!asUnsigned &amp;&amp; negative) buff[count++] = '-';
 buff[count] = 0;
 return reverse(buff, count);
 
}
</code></pre>
<a href="https://godbolt.org/z/MdMGerTTs" rel="nofollow noreferrer">https://godbolt.org/z/MdMGerTTs</a>
<pre><code>int main(void) {
 char str[32];
 printf(&quot;%s\n&quot;, convert(str, -15, 8, 1)); 
 printf(&quot;%s\n&quot;, convert(str, -15, 8, 0)); 

}
</code></pre>
Result:
<pre><code>37777777761
-17
</code></pre>

blocks|key|1369503|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1369504|，我正在实现printf函数，现在只能将负基数10转换为8八进制数。|blockquote|1369505|1369506|1369507|这意味着OP不正确地获得了参数。|1369508|对于"%25o"、"%25x"、"%25X"、"%25u"，对应的参数是unsigned类型，而不是int类型。|offset|length|style|CODE|1369509|对于正确的printf()，不需要打印带说明符oxXu的有符号整数。|1369510|试图传递负整数以匹配oxXu的代码将导致未定义的行为(UB)。常见的UB是将已签名的参数视为未签名的。|1369511|entityMap^0|0|0|0|0|0|2|4|7|4|C|4|H|4|S|8|16|3|0|5|8|N|4|0|A|4|0^^$0|@$1|2|3|-4|4|5|6|T|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|U|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|V|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|W|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|X|7|@]|8|@]|9|$]]|$1|H|3|I|4|5|6|Y|7|@$J|Z|K|10|L|M]|$J|11|K|12|L|M]|$J|13|K|14|L|M]|$J|15|K|16|L|M]|$J|17|K|18|L|M]|$J|19|K|1A|L|M]]|8|@]|9|$]]|$1|N|3|O|4|5|6|1B|7|@$J|1C|K|1D|L|M]|$J|1E|K|1F|L|M]]|8|@]|9|$]]|$1|P|3|Q|4|5|6|1G|7|@$J|1H|K|1I|L|M]]|8|@]|9|$]]|$1|R|3|-4|4|5|6|1J|7|@]|8|@]|9|$]]]|S|$]]

<blockquote>
I'm implementing the printf function and now am stuck at the conversion of the negative base 10 number into a base 8 octal number
</blockquote>
This implies OP is incorrectly getting the argument.
With <code>&quot;%o&quot;</code>, <code>&quot;%x&quot;</code>, <code>&quot;%X&quot;</code>, <code>&quot;%u&quot;</code>, the corresponding argument is type <code>unsigned</code>, not <code>int</code>.
With a proper <code>printf()</code>, there is no printing of a signed integer with specifiers <code>oxXu</code>.
Code attempting to pass a negative integer to match a <code>oxXu</code> results in undefined behavior (UB). A common UB is to treat the signed argument as if it was unsigned.

I'm implementing the printf function and now am stuck at the conversion of the negative base 10 number into a base 8 octal number, below i placed the piece of code which is working just fine in regard to converting a positive decimal number to octal just like the printf
<blockquote>
my ft_printf =&gt; 15 octal =&gt; 017
</blockquote>
<blockquote>
original printf =&gt; 15 octal =&gt; 017
</blockquote>
but regarding the negative numbers:
<blockquote>
my ft_printf =&gt; -15 octal =&gt; 0-17
</blockquote>
<blockquote>
original printf =&gt; -15 octal =&gt; 037777777761
</blockquote>
what should i modify with my function in order for it to print the negative numbers just like the printf does thank you!!!
<pre class="lang-c prettyprint-override"><code>void process_octal(unsigned int n)
{
 char *str;
 int oct = 0;
 int rem = 0;
 int place = 1;

 while (n)
 {
 rem = n % 8;
 oct = oct + rem * place;
 n = n / 8;
 place = place * 10; 
 }
 str = ft_itoa(oct);
 ft_putstr_fd(str, 1);
}
</code></pre>

How to convert a negative decimal number to octal base without any helping funtions

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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我正在实现printf函数，现在我坚持将负基数10转换为8八进制数，在下面我放置了一段代码，它可以很好地将一个正十进制数转换成八进制，就像printf一样。 my ft_printf => 15八进制=> 017原始打印本=> 15八进制=> 017但关于负数： my ft_printf => -15八进制=> 0-1...

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