C语言 如何使用 printf 重复一个字符？
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How to repeat a char using printf?
提问by Muis
I'd like to do something like printf("?", count, char)to repeat a character counttimes.
我想做一些事情，比如printf("?", count, char)重复一个角色的count时间。
What is the right format-string to accomplish this?
完成此操作的正确格式字符串是什么？
EDIT: Yes, it is obvious that I could call printf()in a loop, but that is just what I wanted to avoid.
编辑：是的，很明显我可以printf()在循环中调用，但这正是我想要避免的。
采纳答案by synthesizerpatel
Short answer - yes, long answer: not how you want it.
简短的回答 - 是的，长的回答：不是你想要的。
You can use the %* form of printf, which accepts a variable width. And, if you use '0' as your value to print, combinedwith the right-aligned text that's zero padded on the left..
您可以使用 %* 形式的printf，它接受可变宽度。并且，如果您使用“0”作为打印值，并结合在左侧填充零的右对齐文本。
printf("%0*d\n", 20, 0);
produces:
产生：
00000000000000000000
With my tongue firmly planted in my cheek, I offer up this little horror-show snippet of code.
我的舌头牢牢地贴在我的脸颊上，我提供了这个小小的恐怖表演代码片段。
Some times you just gotta do things badlyto remember why you try so hard the rest of the time.
有时候，你只需要把事情做得很糟糕，就可以记住为什么你在剩下的时间里那么努力。
#include <stdio.h>
int width = 20;
char buf[4096];
void subst(char *s, char from, char to) {
while (*s == from)
*s++ = to;
}
int main() {
sprintf(buf, "%0*d", width, 0);
subst(buf, '0', '-');
printf("%s\n", buf);
return 0;
}
回答by rep_movsd
You can use the following technique:
您可以使用以下技术：
printf("%.*s", 5, "=================");
This will print "====="
It works for me on Visual Studio, no reason it shouldn't work on all C compilers.
这将打印"====="
它在 Visual Studio 上对我有用，没有理由它不适用于所有 C 编译器。
回答by Ariel
If you limit yourself to repeating either a 0 or a space you can do:
如果您限制自己重复 0 或空格，则可以执行以下操作：
For spaces:
对于空间：
printf("%*s", count, "");
For zeros:
对于零：
printf("%0*d", count, 0);
回答by Kristian Gregersen
In c++ you could use std::string to get repeated character
在 C++ 中，您可以使用 std::string 来获取重复字符
printf("%s",std::string(count,char).c_str());
For example:
例如：
printf("%s",std::string(5,'a').c_str());
output:
输出：
aaaaa
回答by Mats Petersson
There is no such thing. You'll have to either write a loop using printfor puts, or write a function that copies the string count times into a new string.
哪有这回事。您必须使用printfor编写一个循环puts，或者编写一个函数来将字符串 count 次复制到一个新字符串中。
回答by Keith Thompson
printfdoesn't do that -- and printfis overkill for printing a single character.
printf不这样做 - 并且printf对于打印单个字符来说太过分了。
char c = '*';
int count = 42;
for (i = 0; i < count; i ++) {
putchar(c);
}
Don't worry about this being inefficient; putchar()buffers its output, so it won't perform a physical output operation for each character unless it needs to.
不要担心这会效率低下；putchar()缓冲其输出，因此除非需要，否则不会对每个字符执行物理输出操作。
回答by ramon
If you have a compiler that supports the alloca() function, then this is possible solution (quite ugly though):
如果您有一个支持 alloca() 函数的编译器，那么这是可能的解决方案（虽然很丑陋）：
printf("%s", (char*)memset(memset(alloca(10), 'for (int i = 0; i < 10; i++, printf("%c", 'x'));
', 10), 'x', 9));
It basically allocates 10 bytes on the stack which are filled with '\0' and then the first 9 bytes are filled with 'x'.
它基本上在堆栈上分配 10 个字节，用 '\0' 填充，然后前 9 个字节用 'x' 填充。
If you have a C99 compiler, then this might be a neater solution:
如果您有 C99 编译器，那么这可能是一个更简洁的解决方案：
#include <stdio.h>
#include <string.h>
void repeat_char(unsigned int cnt, char ch) {
char buffer[cnt + 1];
/*assuming you want to repeat the c character 30 times*/
memset(buffer,ch,cnd); buffer[cnt]='#include <stdio.h>
void repeat (char input , int count )
{
for (int i=0; i != count; i++ )
{
printf("%c", input);
}
}
int main()
{
repeat ('#', 5);
return 0;
}
';
printf("%s",buffer)
}
回答by Pavlos Fragkiadoulakis
#####
回答by Loay Hussein
you can make a function that do this job and use it
你可以制作一个功能来完成这项工作并使用它
printf("%.*s\n",n,(char *) memset(buffer,c,n));
This will output
这将输出
##代码##回答by Rene
n<= sizeof(buffer)[ maybe also n < 2^16]
n<= sizeof(buffer)[ 也可能 n < 2^16]
However the optimizer may change it to puts(buffer)and then the lack of EoS will .....
但是优化器可能会将其更改为puts(buffer)然后缺少 EoS 将.....
And the assumption is that memset is an assembler instruction (but still a loop be it on chip).
并且假设 memset 是一个汇编指令（但在芯片上仍然是一个循环）。
Strictly seen there is no solution given you precondition 'No loop'.
严格来说，给定“无循环”的前提条件，没有解决方案。
