整数到“字符串”转换 [英] Integer to "string" conversions
问题描述
现在，我阅读了faq，它建议使用sprintf。但是，
我想知道字符串中整数的完成位置。
基本上，我想：
nbr |其他数据
但其他数据的所有方式都必须从同一个地方开始。我有一些问题使用sprintf来完成这个要求。
也许
我忽略了一些东西。但是sprintf将nbr完全翻译成
字符串，所以nbr 123最终会占用：
p [0 ] =''1''
p [1] =''2''
p [2] =''3''
随着数量的增加，占用的空间越来越大。
为了解决这个问题，我选择了下面的解决方案。但是这需要
我
以后要退回数字。有没有更好的方法呢？
注意：我使用uint32_t来表示32位无符号整数。这个
是具体的实施
，但为了这个讨论，我需要知道
的大小
$ b $提前分配给buf的整数的b。我也是
忽略
动态分配数组也用于讨论。
#include< stdio.h>
#include< stdlib.h>
#include< string.h>
int
main（void）{
int i;
uint32_t nbr;
unsigned char buf [4];
nbr = 0xffffffff;
memset（buf，0，sizeof buf）;
buf [0] =（ unsigned char）（nbr> 24）& 0xff;
buf [1] =（unsigned char）（nbr> 16）& 0xff;
buf [2] =（unsigned char）（nbr> 8）& 0xff;
buf [3] =（unsigned char）nbr& 0xff;
for（i = 0; i< 4; i ++）
printf（"％d \ nn"，buf [i] ）;
退出（EXIT_SUCCESS）;
}
bw*****@yahoo.com 写道：
现在，我读了faq，它建议使用sprintf。但是，
我想知道字符串中整数的完成位置。
基本上，我想：
nbr |其他数据
但其他数据的所有方式都必须从同一个地方开始。我有一些问题使用sprintf来完成这个要求。
也许
我忽略了一些东西。 [...]
可能类似于％5d的事情或者％05d。
-
Eric Sosman
es ***** @ acm-dot-org.inva lid
bw ***** @ yahoo.com 写道：
现在，我读了faq，它建议使用sprintf。但是，
我想知道字符串中整数的完成位置。
基本上，我想：
nbr |其他数据
但其他数据的所有方式都必须从同一个地方开始。我有一些问题使用sprintf来完成这个要求。
也许
我忽略了一些东西。但是sprintf将nbr完全翻译成
字符串，所以nbr 123最终会占用：
p [0 ] =''1''
p [1] =''2''
p [2] =''3''
随着数量的增加，占用的空间越来越大。
为了解决这个问题，我选择了下面的解决方案。但是这需要
我
以后要退回数字。有没有更好的方法呢？
注意：我使用uint32_t来表示32位无符号整数。这个
是具体的实施
，但为了这个讨论，我需要知道
的大小
$ b $提前分配给buf的整数的b。我也是
忽略
动态分配数组也用于讨论。
#include< stdio.h>
#include< stdlib.h>
#include< string.h>
int
main（void）{
int i;
uint32_t nbr;
unsigned char buf [4];
nbr = 0xffffffff;
memset（buf，0，sizeof buf）;
buf [0] =（ unsigned char）（nbr> 24）& 0xff;
buf [1] =（unsigned char）（nbr> 16）& 0xff;
buf [2] =（unsigned char）（nbr> 8）& 0xff;
buf [3] =（unsigned char）nbr& 0xff;
for（i = 0; i< 4; i ++）
printf（"％d \ nn"，buf [i] ）;
退出（EXIT_SUCCESS）;
}
你的生活远远不够太难了。
A）如果你使用了以下内容：
char buf [25];
unsigned int num = 3784;
sprintf（buf，"％d"，num）;
你可以轻松使用strlen（buf）来计算sprintf（）
结果有多长。
B）为什么不将所有字符串格式组合成一个进程
使用类似的东西：
char buf [100];
unsigned int num = 23405;
unsigned int num2 = 978 ;
char name [] =" Ben Hurr" ;;
sprintf（buf，"％10d％s - ％d \ n"，num，name， num2）;
- mkaras
bw ***** @ yahoo.com 写道：
>
现在，我读了faq，它建议使用sprintf。但是，
我想知道字符串中整数的完成位置。
基本上，我想：
nbr |其他数据
但其他数据的所有方式都必须从同一个地方开始。我有一些问题使用sprintf来完成这个要求。
也许
我忽略了一些东西。但是sprintf将nbr完全翻译成
字符串，所以nbr 123最终会占用：
p [0 ] =''1''
p [1] =''2''
p [2] =''3''
sizeof" 123"是四个，而不是三个。
p [3] =''\ 0''
-
pete
Now, I read the faq, and it suggests using sprintf. However,
I want to all ways know where the integer finishes in the string.
Basically, I want to:
nbr | other data
But the other data all ways has to start at the same place. I had
some problems using sprintf to accomplish this requirement. Maybe
I am overlooking something. But sprintf translates the nbr exactly
into
the string, so the nbr 123, would end up occupying:
p[0] = ''1''
p[1] = ''2''
p[2] = ''3''
So as the number grew, the space taken grew.
To solve this, I went with the solution below. But this would require
me
to OR back the number later. Is there a better way to do this?
Note: I am using uint32_t to signify a 32 bit unsigned integer. This
is implementation
specific, but for the sake of this discussion, I need to know
the size
of the integer being assigned to buf ahead of time. I am also
ignoring
dynamically allocated arrays for this discussion as well.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int
main(void) {
int i;
uint32_t nbr;
unsigned char buf[4];
nbr = 0xffffffff;
memset(buf, 0, sizeof buf);
buf[0] = (unsigned char)(nbr >24) & 0xff;
buf[1] = (unsigned char)(nbr >16) & 0xff;
buf[2] = (unsigned char)(nbr > 8) & 0xff;
buf[3] = (unsigned char) nbr & 0xff;
for(i = 0; i < 4; i++)
printf("%d\n", buf[i]);
exit(EXIT_SUCCESS);
}
bw*****@yahoo.com wrote:
Now, I read the faq, and it suggests using sprintf. However,
I want to all ways know where the integer finishes in the string.
Basically, I want to:
nbr | other data
But the other data all ways has to start at the same place. I had
some problems using sprintf to accomplish this requirement. Maybe
I am overlooking something. [...]Probably things like "%5d" or "%05d".
--
Eric Sosman
es*****@acm-dot-org.invalid
bw*****@yahoo.com wrote:Now, I read the faq, and it suggests using sprintf. However,
I want to all ways know where the integer finishes in the string.
Basically, I want to:
nbr | other data
But the other data all ways has to start at the same place. I had
some problems using sprintf to accomplish this requirement. Maybe
I am overlooking something. But sprintf translates the nbr exactly
into
the string, so the nbr 123, would end up occupying:
p[0] = ''1''
p[1] = ''2''
p[2] = ''3''
So as the number grew, the space taken grew.
To solve this, I went with the solution below. But this would require
me
to OR back the number later. Is there a better way to do this?
Note: I am using uint32_t to signify a 32 bit unsigned integer. This
is implementation
specific, but for the sake of this discussion, I need to know
the size
of the integer being assigned to buf ahead of time. I am also
ignoring
dynamically allocated arrays for this discussion as well.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int
main(void) {
int i;
uint32_t nbr;
unsigned char buf[4];
nbr = 0xffffffff;
memset(buf, 0, sizeof buf);
buf[0] = (unsigned char)(nbr >24) & 0xff;
buf[1] = (unsigned char)(nbr >16) & 0xff;
buf[2] = (unsigned char)(nbr > 8) & 0xff;
buf[3] = (unsigned char) nbr & 0xff;
for(i = 0; i < 4; i++)
printf("%d\n", buf[i]);
exit(EXIT_SUCCESS);
}You are making your life far far too hard.
A) If you you used the following:
char buf[25];
unsigned int num = 3784;
sprintf(buf, "%d", num);
you could easily use strlen(buf) to figure out how long the sprintf()
result is.
B) Why not combine all the string formatting together into one process
using something like:
char buf[100];
unsigned int num = 23405;
unsigned int num2 = 978;
char name[] = "Ben Hurr";
sprintf(buf, "%10d %s - %d\n", num, name, num2);
- mkaras
bw*****@yahoo.com wrote:>
Now, I read the faq, and it suggests using sprintf. However,
I want to all ways know where the integer finishes in the string.
Basically, I want to:
nbr | other data
But the other data all ways has to start at the same place. I had
some problems using sprintf to accomplish this requirement. Maybe
I am overlooking something. But sprintf translates the nbr exactly
into
the string, so the nbr 123, would end up occupying:
p[0] = ''1''
p[1] = ''2''
p[2] = ''3''
sizeof "123" is four, not three.
p[3] = ''\0''
--
pete
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