C++ 用 sprintf 填充
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padding with sprintf
提问by Abruzzo Forte e Gentile
I have a dummy question. I would like to print an integer into a buffer padding with 0 but I cannot sort it out the sprintfformat.
I am trying the following
我有一个假问题。我想用 0 将整数打印到缓冲区填充中，但我无法对sprintf格式进行排序。我正在尝试以下
char buf[31];
int my_val = 324;
sprintf( buf, "%d030", my_val );
hoping to have the following string
希望有以下字符串
"000000000000000000000000000324"
what am I doing wrong? It doesn't mean pad with 0 for a max width of 30 chars?
我究竟做错了什么？这并不意味着用 0 填充最大宽度为 30 个字符？
回答by Seth Robertson
"%030d"is the droid you are looking for
"%030d"是你要找的机器人吗
回答by Sreerac
You got the syntax slightly wrong; The following code produces the desired output:
你的语法有点错误；以下代码产生所需的输出：
char buf[31];
int my_val = 324;
sprintf( buf, "%030d", (int)my_val );
From Wikipedia's Article on Printf:
[...] printf("%2d", 3) results in " 3", while printf("%02d", 3) results in "03".
回答by Nick Meyer
The padding and width come beforethe type specifier:
填充和宽度在类型说明符之前：
sprintf( buf, "%030d", my_val );
回答by Matthew
Try:
尝试：
sprintf( buf, "%030d", my_val );
回答by jbruni
Your precision and width parameters need to go between the '%' and the conversion specifier 'd', not after. In fact all flags do. So if you want a preceeding '+' for positive numbers, use '%+d'.
您的精度和宽度参数需要在 '%' 和转换说明符 'd' 之间，而不是在后面。事实上，所有的标志都可以。因此，如果您想要前面的 '+' 表示正数，请使用 '%+d'。
回答by John
It's %030d, with type-letter at the end.
它是%030d，末尾带有打字机。
回答by Lundin
A fairly effective version that doesn't need any slow library calls:
一个不需要任何缓慢库调用的相当有效的版本：
#include <stdio.h>
void uint_tostr (unsigned int n, size_t buf_size, char dst[buf_size])
{
const size_t str_size = buf_size-1;
for(size_t i=0; i<str_size; i++)
{
size_t index = str_size - i - 1;
dst[index] = n%10 + '0';
n/=10;
}
dst[str_size] = '##代码##';
}
int main (void)
{
unsigned int n = 1234;
char str[6+1];
uint_tostr(n, 6+1, str);
puts(str);
}
This can be optimized further, though it is still probably some hundred times faster than sprintfas is.
这可以进一步优化，尽管它仍然可能比sprintf现在快一百倍。
