如何在Arduino上格式化长添加千位分隔符

Question

我正在Arduino上开发一个项目，它解析来自远程Web API的一些JSON数据，并在16x2 LCD上显示它。

我想格式化一个长解析TextFinder添加千位分隔符（逗号分隔符将罚款）。

简而言之，我该如何编写formatLong函数？

long longToBeFormatted = 32432423; 

formattedLong = formatLong(longToBeFormatted); //How to implement this? 

lcd.print(formattedLong) // formattedLong is a string 

Answer 1

我不确定在Arduino上使用了哪些工具集。有时候库将支持非标准的“千人分组”标志 - 单引号字符是典型的扩展：

printf("%'ld",long_val); 

如果你的库不支持这一点，像下面这样可能会做：

#include <stddef.h> 
#include <string.h> 
#include <limits.h> 
#include <stdio.h> 
#include <assert.h> 

size_t strlcpy(char* dest, char const* src, size_t dest_size); 

size_t format_long(long x, char* buf, size_t bufsize) 
{ 
    // This code assumes 32-bit long, is that the 
    // case on Arduino? Modifying it to be able to 
    // handle 64-bit longs (or to not care) should be 
    // pretty straightforward if that's necessary. 

    char scratch[sizeof("-2,147,483,648")]; 
    char* p = scratch + sizeof(scratch); // Work from end of buffer 
    int neg = (x < 0); 

    // Handle a couple special cases 
    if (x == 0) { 
     return strlcpy(buf, "0", bufsize); 
    } 
    if (x == INT_MIN) { 
     // Lazy way of handling this special case 
     return strlcpy(buf, "-2,147,483,648", bufsize); 
    } 

    // Work with positive values from here on 
    if (x < 0) x = -x; 

    int group_counter = 3; 
    *(--p) = 0; // Null terminate the scratch buffer 

    while (x != 0) { 
     int digit = x % 10; 
     x = x/10; 

     assert(p != &scratch[0]); 
     *(--p) = ""[digit]; 

     if ((x != 0) && (--group_counter == 0)) { 
      assert(p != &scratch[0]); 
      *(--p) = ','; 
      group_counter = 3; 
     } 
    } 

    if (neg) { 
     assert(p != &scratch[0]); 
     *(--p) = '-'; 
    } 
    return strlcpy(buf, p, bufsize); 
} 


/* 
    A non-optimal strlcpy() implementation that helps copying string 
    without danger of buffer overflow. 

    This is provided just in case you don't have an implementation 
    so the code above will actually compile and run. 
*/ 
size_t strlcpy(char* dest, char const* src, size_t dest_size) 
{ 
    size_t len = strlen(src); 

    if (dest_size == 0) { 
     // nothing to copy - just return how long the buffer should be 
     // (note that the return value doens't include the null terminator) 
     return len; 
    } 

    size_t tocopy = (dest_size <= len) ? dest_size-1 : len; 

    memmove(dest, src, tocopy); 
    dest[tocopy] = 0; 

    return len; 
} 

Answer 2

也许不是最好的算法，但这里的一个实例（标准C）：

char* formatLong(long toBeFormatted) 
{ 
    // Get the string representation as is 
    char* buffer = (char*) malloc(sizeof(long)); 
    ltoa(toBeFormatted, buffer, 10); 

    // Calculate how much commas there will be 
    unsigned int buff_length = strlen(buffer); 
    unsigned int num_commas = buff_length/3; 
    unsigned int digits_left = buff_length % 3; 
    if (digits_left == 0) 
    { 
     num_commas--; 
    } 

    // Allocate space for final string representation 
    unsigned int final_length = buff_length + num_commas + 1; 
    char* final = (char*) malloc(final_length); 
    memset(final, 0, final_length); 

    // Parse strings from last to first to count positions 
    int final_pos = final_length - 2; 
    int buff_pos = buff_length - 1; 
    int i = 0; 
    while(final_pos >= 0) 
    { 
     final[final_pos--] = buffer[buff_pos--]; 
     i++; 
     if (i % 3 == 0) 
     { 
      final[final_pos--] = ','; 
     } 
    } 

    // Free obsolete memory and return buffer 
    free(buffer); 
    return final; 
} 

Answer 3

你可以尝试以下方法：

std::string formatLong(long l) { 
    int power = 0; 
    while (l/(pow(1000, power)) { power += 1; } 

    power -= 1; 

    std::stringstream s; 
    while (power) { 
     s << l/pow(1000, power); 
     s << ","; 
     l % (pow(1000, power)); 
     power -= 1; 
    } 
    return s->str(); 
} 

的代码是从臀部射击，但这个想法应该起作用。