C 语言有 setfill() 替代方案吗？

发布于 2024-09-26 05:20:34 字数 266 浏览 7 评论 0原文

在 C++ 中：

int main()
{
    cout << setfill('#') << setw(10) << 5 << endl;

    return 0;
}

输出：

#########5

C 是否有任何 setfill() 替代方案？或者如何在 C 中执行此操作而无需手动创建字符串？

原文

In C++:

int main()
{
    cout << setfill('#') << setw(10) << 5 << endl;

    return 0;
}

Outputs:

#########5

Is there any setfill() alternative for C? Or how to do this in C without manually creating the string?

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享受孤独 2024-10-03 05:20:34

 int x= 5; 
 printf("%010d",x);

将输出：
0000000005

现在，如果您确实想要“#”而不是“0”，则必须在字符串中手动替换它们。

或许：

char buf[11], *sp = buf; 
snprintf(buf, 11, "%10d", x); 
while( (sp = strchr(sp, ' ')) != '\0'){ *sp = '#'; }
puts(buf);

 int x= 5; 
 printf("%010d",x);

will output :
0000000005

Now if you really want '#' instead of '0' you'll have to replace them manually in the string.

Maybe :

char buf[11], *sp = buf; 
snprintf(buf, 11, "%10d", x); 
while( (sp = strchr(sp, ' ')) != '\0'){ *sp = '#'; }
puts(buf);

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硬不硬你别怂 2024-10-03 05:20:34

不，没有直接的替代方案。

但是，如果您有一个行为良好的 snprintf （其行为符合 C99 标准所描述的），则无需创建新字符串即可工作；仅创建 2 个临时 int

#include <stdio.h>

int main(void) {
  int filler = '#'; /* setfill('#') */
  int width = 10;   /* setw(10)     */
  int target = 5;   /* 5            */

  /* ******** */
  int s = snprintf(NULL, 0, "%d", target);
  for (int i = 0; i < width - s; i++) {
    putchar(filler);
  }
  printf("%d\n", target);
  /* ******** */

  return 0;
}

编辑：在 ideone 运行版本。

EDIT2：C99 标准的 snprintf 和 Windows _snprintf 之间的差异（感谢您的链接，Ben）：

原型：int snprintf( char *限制缓冲区, size_t n, const char *限制格式, ...);
原型: int _snprintf(char *buffer, size_t n, const char *format, ...); code>
snprintf 写入不超过 (n-1) 个字节，NUL
_snprintf 写入不超过 (n) 个字节，最后一个可能是 NUL 或其他字符
snprintf 返回格式所需的字符数（可以大于n）或-1，以防编码错误
_snprintf<如果 n 对于字符串来说不大，/code> 返回负值；或 n 如果 NUL 字节尚未写入缓冲区。

您可以在循环中运行行为不当的 _snprintf，增加 n 直到找到正确的值

/* absolutely not tested, written directly on SO text editor */
int i;
size_t rightvalue = 0;
char buffer[SOME_DEFAULT_VALUE];
do {
    if (sizeof buffer < rightvalue) /* OOPS, BIG BIG OOPS */;
    i = _snprintf(buffer, rightvalue, "%d", 42);
} while (i != rightvalue++);
/* rightvalue already has space for the terminating NUL */

No, there is no direct alternative.

But if you have a well-behaved snprintf (one that behaves as described by the C99 Standard), this works without creating a new string; creating only 2 temporary ints

#include <stdio.h>

int main(void) {
  int filler = '#'; /* setfill('#') */
  int width = 10;   /* setw(10)     */
  int target = 5;   /* 5            */

  /* ******** */
  int s = snprintf(NULL, 0, "%d", target);
  for (int i = 0; i < width - s; i++) {
    putchar(filler);
  }
  printf("%d\n", target);
  /* ******** */

  return 0;
}

EDIT: running version at ideone.

EDIT2: Differences between the C99 Standard's snprintf and Windows _snprintf (thank you for the link, Ben):

prototype: int snprintf(char *restrict buffer, size_t n, const char *restrict format, ...);
prototype: int _snprintf(char *buffer, size_t n, const char *format, ...);
snprintf writes no more than (n-1) bytes and a NUL
_snprintf writes no more than (n) bytes, the last of which may be NUL or other character
snprintf returns the number of characters needed for format (can be larger than n) or -1 in case of encoding error
_snprintf returns a negative value if n is not large for the string; or n if a NUL byte hasn't been written to buffer.

You can run the mis-behaving _snprintf in a loop, increasing n until you find the right value

/* absolutely not tested, written directly on SO text editor */
int i;
size_t rightvalue = 0;
char buffer[SOME_DEFAULT_VALUE];
do {
    if (sizeof buffer < rightvalue) /* OOPS, BIG BIG OOPS */;
    i = _snprintf(buffer, rightvalue, "%d", 42);
} while (i != rightvalue++);
/* rightvalue already has space for the terminating NUL */

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柠檬 2024-10-03 05:20:34

每个人都想调用 printf 两次... printf 是最昂贵的函数之一。

这里：

char buffer[18] = "##########";
puts(buffer + snprintf(buffer+strlen(buffer), 8, "%d", 5));

Everybody wants to call printf twice... printf is one of the most expensive functions around.

Here:

char buffer[18] = "##########";
puts(buffer + snprintf(buffer+strlen(buffer), 8, "%d", 5));

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攒一口袋星星 2024-10-03 05:20:34

不是按照标准内置的，

我可能会 sprintf() 将数字转换为字符串并获取字符数，然后在打印字符串之前首先输出正确的“#”数量。

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丑丑阿 2024-10-03 05:20:34

下面将使用 C 中的 memset 来完成此操作（假设为 1 字节字符）。它仍然创建一个“字符串”，尽管我不确定您不希望它是多么手动。

int main(void)
{

   char buf[MAX_LENGTH];

   memset(buf, '#', 10);
   buf[10]='\0';
   printf("%s5\n", buf);
}

根据您实际想要用它做什么，您可以动态创建适当大小的缓冲区，并根据需要从辅助函数返回它。

The following will do it using memset in C assuming a 1-byte char. It still creates a 'string', though I'm not sure how manually you don't want it to be.

int main(void)
{

   char buf[MAX_LENGTH];

   memset(buf, '#', 10);
   buf[10]='\0';
   printf("%s5\n", buf);
}

Depending what you want to actually do with it, you could dynamically create the buffer to be the appropriate size and return it from a helper function if you so desire.

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