blocks|key|114178|text|我的快速解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|114179|#include+<stdio.h>

static+const+unsigned+char+S[]+=+{0xF0,+0x80,+0x80,+0xF0,+0x10,+0x10,+0xF0};

int+main(void)+{
++++
++++for(int+i=0;+i<sizeof(S);+%2B%2Bi)
++++{
++++++++for(int+b=7;b;+--b)
++++++++{
++++++++++++printf("%25c",+"*+"[!(S[i]+&+(1<<b))]);
++++++++}
++++++++for(int+b=7;b;+--b)
++++++++{
++++++++++++printf("%25c",+"*+"[!(S[i]+&+(1<<b))]);
++++++++}
++++++++printf("\n");
++++}
++++return+0;
}|code-block|syntax|javascript|114180|输出：|114181|Success+#stdin+#stdout+0s+5548KB
****+++****+++
*++++++*++++++
*++++++*++++++
****+++****+++
+++*++++++*+++
+++*++++++*+++
****+++****+++|114182|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

My quickie solution:
<pre><code>#include &lt;stdio.h&gt;

static const unsigned char S[] = {0xF0, 0x80, 0x80, 0xF0, 0x10, 0x10, 0xF0};

int main(void) {
 
 for(int i=0; i&lt;sizeof(S); ++i)
 {
 for(int b=7;b; --b)
 {
 printf(&quot;%c&quot;, &quot;* &quot;[!(S[i] &amp; (1&lt;&lt;b))]);
 }
 for(int b=7;b; --b)
 {
 printf(&quot;%c&quot;, &quot;* &quot;[!(S[i] &amp; (1&lt;&lt;b))]);
 }
 printf(&quot;\n&quot;);
 }
 return 0;
}
</code></pre>
<h2>Output:</h2>
<pre><code>Success #stdin #stdout 0s 5548KB
**** **** 
* * 
* * 
**** **** 
 * * 
 * * 
**** **** 
</code></pre>

blocks|key|114234|text|在处理这样的问题时，最重要的是将代码分解成更小的部分，使用函数和/或类型防御或结构来组织事情。在您的情况下，您可以很容易地使用函数将如何打印信函的星星和空格的“知识”与主代码(该代码知道要打印哪个字母的顺序)分开。通过这种方式重构代码，我们可以得到：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|114235|void+S(int+k,+int+l)+{
+++if(k==1+&&+l<=4){
++++++++printf("*");
++++}
++++else+if(k>=2+&&+k<=3+&&+l==1){
++++++++printf("*");
++++}
++++else+if(k==4+&&+l<=4){
++++++++printf("*");
++++}
++++else+if(k>=5+&&+k<=6+&&+l==4){
++++++++printf("*");
++++}
++++else+if(k==7+&&+l<=4){
++++++++printf("*");
++++}
++++else{
++++++++printf("+");
++++}
}

int+main()+{
++++int+k,l,i,j;
++++for(k=1;k<=7;k%2B%2B){
++++++++for(l=1;l<=4;l%2B%2B){
++++++++++++S(k,+l);
++++++++}
++++++++printf("\n");
++++}
++++for(i=1;i<=7;i%2B%2B){
++++++++for(j=1;j<=4;j%2B%2B){
++++++++++++S(i,+j);
++++++++}
++++++++printf("\n");
++++}
++++return+0;
}|code-block|syntax|javascript|114236|现在，您可以轻松地重写main以并排打印字母。|114237|int+main()+{
++++int+k,l,i,j;
++++for(k=1;k<=7;k%2B%2B){
++++++++for(l=1;l<=4;l%2B%2B){
++++++++++++S(k,+l);
++++++++}
++++++++printf("+");
++++++++for(j=1;j<=4;j%2B%2B){
++++++++++++S(k,+j);
++++++++}
++++++++printf("\n");
++++}
++++return+0;
}|114238|您也可以通过以不同的模式书写不同的字母来扩展它(虽然每个字母的功能有点冗长--最好是将所有字母的像素存储在一个数据结构中并以这种方式打印出来)。|114239|entityMap^0|0|0|0|0|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]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

The most important thing when approaching problems like this is to decompose your code in smaller pieces, using functions and/or typedefs or structs to organize things. In your case, you can easily separate the &quot;knowledge&quot; of how to print stars and spaces for your letters from the main code (which knows which letters to print in which order) using functions. Refactoring you code that way, we get:
<pre><code>void S(int k, int l) {
 if(k==1 &amp;&amp; l&lt;=4){
 printf(&quot;*&quot;);
 }
 else if(k&gt;=2 &amp;&amp; k&lt;=3 &amp;&amp; l==1){
 printf(&quot;*&quot;);
 }
 else if(k==4 &amp;&amp; l&lt;=4){
 printf(&quot;*&quot;);
 }
 else if(k&gt;=5 &amp;&amp; k&lt;=6 &amp;&amp; l==4){
 printf(&quot;*&quot;);
 }
 else if(k==7 &amp;&amp; l&lt;=4){
 printf(&quot;*&quot;);
 }
 else{
 printf(&quot; &quot;);
 }
}

int main() {
 int k,l,i,j;
 for(k=1;k&lt;=7;k++){
 for(l=1;l&lt;=4;l++){
 S(k, l);
 }
 printf(&quot;\n&quot;);
 }
 for(i=1;i&lt;=7;i++){
 for(j=1;j&lt;=4;j++){
 S(i, j);
 }
 printf(&quot;\n&quot;);
 }
 return 0;
}
</code></pre>
Now you can easily rewrite main to print the letters side by side
<pre><code>int main() {
 int k,l,i,j;
 for(k=1;k&lt;=7;k++){
 for(l=1;l&lt;=4;l++){
 S(k, l);
 }
 printf(&quot; &quot;);
 for(j=1;j&lt;=4;j++){
 S(k, j);
 }
 printf(&quot;\n&quot;);
 }
 return 0;
}
</code></pre>
You could also extend this by writing different letters in different patterns (though having a function for each letter is somewhat verbose -- better would be to store the pixels for all letters in a data structure and print it that way)

blocks|key|114260|text|对要打印的每个字符使用模式矩阵。这样印刷就更容易了。您将遍历字符串的每个字母，以便为模式中的每一行打印。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|114261|#include+<stdio.h>
#include+<string.h>

#define+ROWMAX+7
#define+COLMAX+5
#define+CSPACE+1
#define+PATTERN+'#'


char+cCube+[26][ROWMAX][COLMAX]+=+{
++++{+++{0,+1,+1,+1,+1},+//+letter+S
++++++++{1,+1,+0,+0,+1},
++++++++{1,+1,+0,+0,+0},
++++++++{0,+1,+1,+1,+0},
++++++++{0,+0,+0,+1,+1},
++++++++{1,+0,+0,+1,+1},
++++++++{1,+1,+1,+1,+0}
++++},
++++{+++{0,+1,+1,+1,+0},+//+letter+O
++++++++{1,+1,+0,+1,+1},
++++++++{1,+0,+0,+0,+1},
++++++++{1,+0,+0,+0,+1},
++++++++{1,+0,+0,+0,+1},
++++++++{1,+1,+0,+1,+1},
++++++++{0,+1,+1,+1,+0}
++++},
++++{+++{1,+1,+1,+1,+0},+//+letter+D
++++++++{1,+0,+0,+1,+1},
++++++++{1,+0,+0,+0,+1},
++++++++{1,+0,+0,+0,+1},
++++++++{1,+0,+0,+0,+1},
++++++++{1,+0,+0,+1,+1},
++++++++{1,+1,+1,+1,+0}
++++},+//so+on+for+alphabets+you+want+to+use
};


int+chIndex+[256]+=+{0};

void+print_pat_msg+(const+char*+msg)+{
++++int+mlen+=+strlen+(msg);
++++for+(int+ri+=+0;+ri+<+ROWMAX;+%2B%2Bri)+{
++++++++for+(int+mi+=+0;+mi+<+mlen;+%2B%2Bmi)+{
++++++++++++int+chi+=+chIndex[+(unsigned+char)+msg[mi]];
++++++++++++for+(int+ci+=+0;+ci+<+COLMAX;+%2B%2Bci)
++++++++++++++++(cCube[chi][ri][ci])+?+putchar+(PATTERN)+:+putchar+('+');
++++++++++++for+(int+spi+=+0;+spi%2B%2B+<+CSPACE;)+putchar+('+');
++++++++}
++++++++putchar+('\n');
++++}
++++printf+("\n");
}

int+main+()+{
++++//+if+patterns+are+indexed+alphabetically+below+initialisation+can+be+omitted
++++chIndex[+(unsigned+char)+'S']+=+0;
++++chIndex[+(unsigned+char)+'O']+=+1;
++++chIndex[+(unsigned+char)+'D']+=+2;

++++char+message+[]+=+"ODDS";
++++print_pat_msg+(message);
++++return+0;
}|code-block|syntax|javascript|114262|114263|通过这种方式，您可以在一个地方轻松地设计/定制每个字母的图案。代码和模式可以在其他消息中重用&+names.|unordered-list-item|114264|Printing在这里的工作方式有点像点阵打印机，在字符串中的每个字母上一次划出一个图案行。|114265|您也可以用几乎任何字符替换0/1标志来绘制ASCII图像。您只需调整模式数组，这很容易跟踪和随意更改。|offset|length|style|CODE|114266|开始思考，这是字体在像素级的工作方式，模式。|114267|。|114268|entityMap^0|0|0|0|0|0|D|1|F|1|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|10|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|11|8|@]|9|@]|A|$]]|$1|K|3|L|5|J|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|J|7|13|8|@$O|14|P|15|Q|R]|$O|16|P|17|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|J|7|18|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|19|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|X|$]]

Use pattern matrix for every character you want to print. It'll make printing easier. You run through each letter of the string to print for every row in the pattern.
<pre class="lang-c prettyprint-override"><code>#include &lt;stdio.h&gt;
#include &lt;string.h&gt;

#define ROWMAX 7
#define COLMAX 5
#define CSPACE 1
#define PATTERN '#'


char cCube [26][ROWMAX][COLMAX] = {
 { {0, 1, 1, 1, 1}, // letter S
 {1, 1, 0, 0, 1},
 {1, 1, 0, 0, 0},
 {0, 1, 1, 1, 0},
 {0, 0, 0, 1, 1},
 {1, 0, 0, 1, 1},
 {1, 1, 1, 1, 0}
 },
 { {0, 1, 1, 1, 0}, // letter O
 {1, 1, 0, 1, 1},
 {1, 0, 0, 0, 1},
 {1, 0, 0, 0, 1},
 {1, 0, 0, 0, 1},
 {1, 1, 0, 1, 1},
 {0, 1, 1, 1, 0}
 },
 { {1, 1, 1, 1, 0}, // letter D
 {1, 0, 0, 1, 1},
 {1, 0, 0, 0, 1},
 {1, 0, 0, 0, 1},
 {1, 0, 0, 0, 1},
 {1, 0, 0, 1, 1},
 {1, 1, 1, 1, 0}
 }, //so on for alphabets you want to use
};


int chIndex [256] = {0};

void print_pat_msg (const char* msg) {
 int mlen = strlen (msg);
 for (int ri = 0; ri &lt; ROWMAX; ++ri) {
 for (int mi = 0; mi &lt; mlen; ++mi) {
 int chi = chIndex[ (unsigned char) msg[mi]];
 for (int ci = 0; ci &lt; COLMAX; ++ci)
 (cCube[chi][ri][ci]) ? putchar (PATTERN) : putchar (' ');
 for (int spi = 0; spi++ &lt; CSPACE;) putchar (' ');
 }
 putchar ('\n');
 }
 printf (&quot;\n&quot;);
}

int main () {
 // if patterns are indexed alphabetically below initialisation can be omitted
 chIndex[ (unsigned char) 'S'] = 0;
 chIndex[ (unsigned char) 'O'] = 1;
 chIndex[ (unsigned char) 'D'] = 2;

 char message [] = &quot;ODDS&quot;;
 print_pat_msg (message);
 return 0;
}
</code></pre>
<ul>
<li>This way you can design/customise the pattern for each letter with ease at one place. Code &amp; pattern can be reused for other messages &amp; names.</li>
<li>Printing here works somewhat like a dot-matrix printer, striping across putting out pattern-line of each letter in the string one at a time.</li>
<li>You can also replace <code>0</code>/<code>1</code> flags with pretty much any character to draw ASCII images. You just have to adjust the pattern array, which is easy to follow &amp; change at will.</li>
<li>Come to think of it, this is how fonts work at pixel level, patterns.</li>
</ul>

I am trying to print the initials of My First Name &quot;S S&quot; using * pattern in C. However the patterns are printed one after the other. Is there any way to print the alphabets using * pattern one at the side of the other with a little space in between. I do not want to allot a very big box for both the alphabets. I want to code for individual alphabets and put them side by side with some space in between. My Code and out our is given below:
<pre><code>int k,l,i,j;
for(k=1;k&lt;=7;k++){
 for(l=1;l&lt;=4;l++){
 if(k==1 &amp;&amp; l&lt;=4){
 printf(&quot;*&quot;);
 }
 else if(k&gt;=2 &amp;&amp; k&lt;=3 &amp;&amp; l==1){
 printf(&quot;*&quot;);
 }
 else if(k==4 &amp;&amp; l&lt;=4){
 printf(&quot;*&quot;);
 }
 else if(k&gt;=5 &amp;&amp; k&lt;=6 &amp;&amp; l==4){
 printf(&quot;*&quot;);
 }
 else if(k==7 &amp;&amp; l&lt;=4){
 printf(&quot;*&quot;);
 }
 else{
 printf(&quot; &quot;);
 }
 }
 printf(&quot;\n&quot;);
}
for(i=1;i&lt;=7;i++){
 for(j=1;j&lt;=4;j++){
 if(i==1 &amp;&amp; j&lt;=4){
 printf(&quot;*&quot;);
 }
 else if(i&gt;=2 &amp;&amp; i&lt;=3 &amp;&amp; j==1){
 printf(&quot;*&quot;);
 }
 else if(i==4 &amp;&amp; j&lt;=4){
 printf(&quot;*&quot;);
 }
 else if(i&gt;=5 &amp;&amp; i&lt;=6 &amp;&amp; j==4){
 printf(&quot;*&quot;);
 }
 else if(i==7 &amp;&amp; j&lt;=4){
 printf(&quot;*&quot;);
 }
 else{
 printf(&quot; &quot;);
 }
 }
 printf(&quot;\n&quot;);
}
return 0;
}
</code></pre>
Output:
<pre><code>****
*
*
****
 *
 *
****
****
*
*
****
 *
 *
****
</code></pre>
However, I want my expected output to be:
<pre><code>**** ****
* *
* *
**** ****
 * *
 * *
**** ****
</code></pre>

printing alphabets using star pattern side by side 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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我试图用C中的*模式来打印我的名字" S“的首字母，但是这些图案是一个接一个地打印出来的。有没有办法用*模式打印字母表，其中一个在另一边，中间有一点空格。我不想为这两个字母分配一个很大的盒子。我想为个人的字母编码，并把它们并排在中间的一些空格。“我的守则”及“我们的守则”如下：int k,l,i,j;for(k=1;k...

问在C中使用星型并排打印字母
EN

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