blocks|key|3922435|text|您可以通过两个步骤完成此操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3922436|#include+<stdio.h>

int+main(void)+{
++++int+var1+=+1048575;
++++int+var2+=+1048575;
++++int+length+=+0;
++++char+buffer[100];
++++char+paddedbuffer[100]+=+"000000000000000";
++++length+=+sprintf+(buffer,+"%25x%25x",+var1,+var2);
++++char*+here+=+length+<+16+?+paddedbuffer+%2B+16+-+length+:+paddedbuffer;
++++sprintf(here,+"%25s",+buffer);
++++printf("%25s\n",+paddedbuffer);
++++return+0;
}|code-block|syntax|javascript|3922437|在某些地方可能有更好的解决方案。|3922438|另一种可能是将结果转换为整数，并使用字段宽度对其进行sprintf，或者首先计算最右侧数字的实际位数。|offset|length|style|CODE|3922439|entityMap^0|0|0|0|Q|7|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@$K|U|L|V|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

You could do it in two steps:

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

int main(void) {
 int var1 = 1048575;
 int var2 = 1048575;
 int length = 0;
 char buffer[100];
 char paddedbuffer[100] = "000000000000000";
 length = sprintf (buffer, "%x%x", var1, var2);
 char* here = length &lt; 16 ? paddedbuffer + 16 - length : paddedbuffer;
 sprintf(here, "%s", buffer);
 printf("%s\n", paddedbuffer);
 return 0;
}
</code></pre>

There's probably a more elegant solution somewhere.

Another possibility is converting your result into an integer and <code>sprintf</code>ing that with a field width, or calculating the actual digits of the rightmost number first.

blocks|key|3922441|text|#include+<stdio.h>

int+main+()
{
++++int+var1=1048575;
++++int+var2=1048575;

++++char+buffer+[17];
++++char+zeros+[]+=+"0000000000000000";++++
++++sprintf+(buffer,+"%25x%25x",+var1,+var2);

++++//+Put+the+zeros+in+front
++++int+len+=+strlen(buffer);
++++if+(len+<+16)
++++{
++++++++zeros[16-len]+=+0x0;+//+Terminate+the+string+at+16-len
++++}
++++else
++++{
++++++++zeros[0]+=+0x0;++//+Terminate+the+string+at+0+as+no+extra+zero+is+needed
++++}

++++printf("%25s%25s\n",+zeros,+buffer);

++++return+0;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3922442|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;

int main ()
{
 int var1=1048575;
 int var2=1048575;

 char buffer [17];
 char zeros [] = "0000000000000000"; 
 sprintf (buffer, "%x%x", var1, var2);

 // Put the zeros in front
 int len = strlen(buffer);
 if (len &lt; 16)
 {
 zeros[16-len] = 0x0; // Terminate the string at 16-len
 }
 else
 {
 zeros[0] = 0x0; // Terminate the string at 0 as no extra zero is needed
 }

 printf("%s%s\n", zeros, buffer);

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

blocks|key|142287|text|您还可以进行一些小运算来计算有效位数，然后打印所需的前导零的数量，因为log_base_+16(+var+)是var的十六进制大小：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|142288|#include+<stdio.h>
#include+<math.h>

int+main+()
{
++++int+var1+=+1048575;
++++int+var2+=+1048575;
++++int+padd+=+16+-+ceil(log(var1)/log(16))-ceil(log(var2)/log(16));
++++char+buffer+[100];
++++sprintf+(buffer,+"%250*x%25x%25x",+padd,+0,+var1,+var2);
++++printf("%25s\n",+buffer);

++++return+0;
}|code-block|syntax|javascript|142289|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|$]]

You can also do some small calculus to compute the number of significant digits and then print the desired number of leading zeros as log_base_ 16(var) is the size of var in hexadecimal:

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

int main ()
{
 int var1 = 1048575;
 int var2 = 1048575;
 int padd = 16 - ceil(log(var1)/log(16))-ceil(log(var2)/log(16));
 char buffer [100];
 sprintf (buffer, "%0*x%x%x", padd, 0, var1, var2);
 printf("%s\n", buffer);

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

blocks|key|2576189|text|使用临时缓冲区和'0'字符串(字符数组)的另一个小技巧是：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2576190|#include+<stdio.h>
#include+<string.h>

enum+{+PADLN+=+16,+MAXC+=+17+};

int+main+(void)
{
++++int+var1=1048575;
++++int+var2=1048575;

++++char+tmp+[MAXC];
++++char+buffer+[MAXC];
++++char+zeros[]+=+"0000000000000000";
++++size_t+len+=+0;

++++sprintf+(tmp,+"%25x%25x",+var1,+var2);
++++len+=+strlen(tmp);
++++zeros[PADLN+-+(len+<+PADLN+?+len+:+PADLN)]+=+0;
++++sprintf+(buffer,+"%25s%25s",+zeros,+tmp);

++++printf+("\n+buffer+:+'%25s'\n\n",+buffer);

++++return+0;
}|code-block|syntax|javascript|2576191|输出|BOLD|2576192|$+./bin/pad16

+buffer+:+'000000ffffffffff'|2576193|entityMap^0|8|3|0|0|0|2|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|U|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|V|8|@$9|W|A|X|B|M]]|D|@]|E|$]]|$1|N|3|O|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|P|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|Q|$]]

Another slight twist using a tmp buffer and a string (char array) of <code>'0'</code>s:

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

enum { PADLN = 16, MAXC = 17 };

int main (void)
{
 int var1=1048575;
 int var2=1048575;

 char tmp [MAXC];
 char buffer [MAXC];
 char zeros[] = "0000000000000000";
 size_t len = 0;

 sprintf (tmp, "%x%x", var1, var2);
 len = strlen(tmp);
 zeros[PADLN - (len &lt; PADLN ? len : PADLN)] = 0;
 sprintf (buffer, "%s%s", zeros, tmp);

 printf ("\n buffer : '%s'\n\n", buffer);

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

Output

<pre><code>$ ./bin/pad16

 buffer : '000000ffffffffff'
</code></pre>

blocks|key|672063|text|打印var2，然后打印连接|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|672064|#include+<stdio.h>

int+main(void)+{
++int+var1+=+1048575;
++int+var2+=+1048575;

++char+buffer[100];
++char+bufvar2[sizeof+var2+*+CHAR_BIT+%2B+1];
++int+len+=+sprintf(bufvar2,+"%25x",+var2);

++#define+TOTAL_MIN_WIDTH+16
++sprintf(buffer,+"%250*x%25s",+TOTAL_MIN_WIDTH+-+len,+var1,+bufvar2);
++printf("%25s\n",+buffer);

++return+0;
}|code-block|syntax|javascript|672065|entityMap^0|2|4|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|$]]

Print <code>var2</code> and then print the concatenation

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

int main(void) {
 int var1 = 1048575;
 int var2 = 1048575;

 char buffer[100];
 char bufvar2[sizeof var2 * CHAR_BIT + 1];
 int len = sprintf(bufvar2, "%x", var2);

 #define TOTAL_MIN_WIDTH 16
 sprintf(buffer, "%0*x%s", TOTAL_MIN_WIDTH - len, var1, bufvar2);
 printf("%s\n", buffer);

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

blocks|key|663734|text|这是一个没有中间缓冲区的简单解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|663735|#include+<stdio.h>

int+main(void)+{
++++int+var1+=+1048575;
++++int+var2+=+1048575;

++++printf("%250*x%25x\n",+16+-+snprintf(NULL,+0,+"%25x",+var2),+var1,+var2);

++++return+0;
}|code-block|syntax|javascript|663736|有一种特殊的情况，您可能想要定义确切的行为:如果var2+==+0呢？上面的代码将打印一个尾随的0。如果只想打印有效数字，因此没有零值的数字，则可以按以下方式修改语句：|offset|length|style|CODE|663737|++++printf("%250*x%25x\n",+16+-+snprintf(NULL,+0,+"%25.x",+var2),+var1,+var2);|663738|一个小区别:在%25和x之间添加一个.，以指定精度字段0，并将最小位数指定为none。|663739|entityMap^0|0|0|O|9|1C|1|0|0|7|1|9|1|G|1|P|1|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]|$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@$I|11|J|12|K|L]|$I|13|J|14|K|L]|$I|15|J|16|K|L]|$I|17|J|18|K|L]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|R|$]]

Here is a simple solution without an intermediary buffer:

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

int main(void) {
 int var1 = 1048575;
 int var2 = 1048575;

 printf("%0*x%x\n", 16 - snprintf(NULL, 0, "%x", var2), var1, var2);

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

There is a special case for which you might want to define the precise behavior: what if <code>var2 == 0</code>? The above code will print a trailing <code>0</code>. If you want only significant digits printed, and therefore no digits for a zero value, you can modify the statement this way:

<pre><code> printf("%0*x%x\n", 16 - snprintf(NULL, 0, "%.x", var2), var1, var2);
</code></pre>

A small difference: adding a <code>.</code> between <code>%</code> and <code>x</code> to specify a precision field of <code>0</code> specifying the minimum number of digits as none.

The code below prints <code>ffffffffff</code>.

I need to the output to be 16 digits long <code>000000ffffffffff</code> with leading zeros. The <code>var1</code> and <code>var2</code> variables can be different so it is not necessary that I want to pad 6 zeros only. I just need to output to be 16 digits with leading zeros. How should I proceed?

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

int main() {
 int var1 = 1048575;
 int var2 = 1048575;

 char buffer[100];

 sprintf(buffer, "%x%x", var1, var2);
 printf("%s\n", buffer);

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

sprintf leading zero padding in C

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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下面的代码打印ffffffffff。我需要的输出是16位数的长度与前导零000000ffffffffff。var1和var2变量可以是不同的，所以我没有必要只填充6个零。我只需要输出到带有前导零的16位数字。我应该如何继续？#include <stdio.h>int main() { int var1 = 104...

问sprintf C++中的前导零填充
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问sprintf C++中的前导零填充EN