C 格式的字符串 - 如何使用 sprintf 向字符串值添加前导零? [英] C formatted string - How to add leading zeros to string value using sprintf?

问题描述

我想在字符串的开头添加零.我正在使用格式说明符.我的输入字符串是 hello我希望输出为 000hello.

我知道如何对整数执行此操作.

int main(){国际我= 232;字符 str[21];sprintf(str,"%08d",i);printf("%s",str);返回0；}

<块引用>

输出将为 -- 00000232

如果我对字符串也这样做.

 int main(){char i[]="你好";字符 str[21];sprintf(str,"%08s",i);printf("%s",str);返回0；}

<块引用>

OUTPUT 将是 - hello(前导 3 个空格)

为什么在字符串的情况下给空间而在整数的情况下给零?

解决方案

如何使用 sprintf 向字符串值添加前导零?

使用 "%0*d%s" 在前面加上零.

"%0*d" --> 0 零的最小宽度，* 从参数列表导出的宽度，d 打印一个 int.

当字符串前面不需要零时，需要一个异常.

void PrependZeros(char *dest, const char *src, unsigned width) {size_t len = strlen(src);if (len >= width) strcpy(dest, src);else sprintf(dest, "%0*d%s", (int) (width - len), 0, src);}

<小时>

但我不认为 sprintf() 是适合这项工作的工具，代码如下.

//根据需要在前面加上0"，从而产生一个 _minimal_ 宽度的字符串.void PrependZeros(char *dest, const char *src, unsigned minimum_width) {size_t len = strlen(src);size_t zeros = (len > minimum_width) ?0 : minimum_width - len;memset(dest, '0', 零);strcpy(dest + zeros, src);}

<小时>

void testw(const char *src, unsigned width) {字符 dest[100];PrependZeros(dest, src, width);printf("%u <%s>\n", width, dest);}int main() {for (无符号 w = 0; w <10; w++)testw("你好", w);for (无符号 w = 0; w <2; w++)testw("", w);}

输出

0 <你好>1 <你好>2 <你好>3 <你好>4 <你好>5 <你好>6 <0你好>7 <00你好>8 <000你好>9 <0000你好>0 <>1<0>

I want to add zeroes at the starting of a string. I am using format specifier. My input string is hello I want output as 000hello.

I know how to do this for integer.

int main()
{
    int i=232;
    char str[21];
    sprintf(str,"%08d",i);

    printf("%s",str);

    return 0;
}

OUTPUT will be -- 00000232

If I do the same for string.

 int main()
    {
        char i[]="hello";
        char str[21];
        sprintf(str,"%08s",i);

        printf("%s",str);

        return 0;
    }

OUTPUT will be - hello (with 3 leading space)

Why it is giving space in case of string and zero in case of integer?

解决方案

How to add leading zeros to string value using sprintf?

Use "%0*d%s" to prepend zeros.

"%0*d" --> 0 min width of zeros, * derived width from the argument list, d print an int.

An exception is needed when the string needs no zeros up front.

void PrependZeros(char *dest, const char *src, unsigned width) {
  size_t len = strlen(src);
  if (len >= width) strcpy(dest, src);
  else sprintf(dest, "%0*d%s", (int) (width - len), 0, src);
}

---

Yet I do not think sprintf() is the right tool for the job and would code as below.

// prepend "0" as needed resulting in a string of _minimal_ width.
void PrependZeros(char *dest, const char *src, unsigned minimal_width) {
  size_t len = strlen(src);
  size_t zeros = (len > minimal_width) ? 0 : minimal_width - len;
  memset(dest, '0', zeros);
  strcpy(dest + zeros, src);
}

---

void testw(const char *src, unsigned width) {
  char dest[100];
  PrependZeros(dest, src, width);
  printf("%u <%s>\n", width, dest);
}

int main() {
  for (unsigned w = 0; w < 10; w++)
    testw("Hello", w);
  for (unsigned w = 0; w < 2; w++)
    testw("", w);
}

Output

0 <Hello>
1 <Hello>
2 <Hello>
3 <Hello>
4 <Hello>
5 <Hello>
6 <0Hello>
7 <00Hello>
8 <000Hello>
9 <0000Hello>
0 <>
1 <0>

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