最小化sprintf或printf的实现 [英] Minimal implementation of sprintf or printf
问题描述
我正在处理一个嵌入式DSP，速度至关重要，内存很短。
I'm working on an embedded DSP where speed is crucial, and memory is very short.
目前，sprintf使用了大部分功能的资源我的代码我只使用它来格式化一些简单的文本：％d，％e，％f，％s ，没有精确或异国情调的操作。
At the moment, sprintf uses the most resources of any function in my code. I only use it to format some simple text: %d, %e, %f, %s, nothing with precision or exotic manipulations.
如何实现一个更适合我使用的基本sprintf或printf函数？
How can I implement a basic sprintf or printf function that would be more suitable for my usage?
推荐答案
这个假设存在一个 itoa 将int转换为字符表示，并且 fputs 写出一个字符串到任何你想要的地方。
This one assumes the existence of an itoa to convert an int to character representation, and an fputs to write out a string to wherever you want it to go.
浮点输出至少在一个方面是不符合的：它不会正确舍入，因为标准要求，所以如果您有（例如）$ 1.234 的值被内部存储为 1.2399999774 ，那么打印为 1.2399 而不是 1.2340 。这节省了相当多的工作，并且对于大多数典型的目的而言仍然足够。
The floating point output is non-conforming in at least one respect: it makes no attempt at rounding correctly, as the standard requires, so if you have have (for example) a value of 1.234 that is internally stored as 1.2399999774, it'll be printed out as 1.2399 instead of 1.2340. This saves quite a bit of work, and remains sufficient for most typical purposes.
这也支持％c 和％x 除了您询问的转化，但如果您想要摆脱这些转换，它们很简单，删除它们（这样做显然会保存一个
This also supports %c and %x in addition to the conversions you asked about, but they're pretty trivial to remove if you want to get rid of them (and doing so will obviously save a little memory).
#include <stdarg.h>
#include <stdio.h>
#include <string.h>
#include <windows.h>
static void ftoa_fixed(char *buffer, double value);
static void ftoa_sci(char *buffer, double value);
int my_vfprintf(FILE *file, char const *fmt, va_list arg) {
int int_temp;
char char_temp;
char *string_temp;
double double_temp;
char ch;
int length = 0;
char buffer[512];
while ( ch = *fmt++) {
if ( '%' == ch ) {
switch (ch = *fmt++) {
/* %% - print out a single % */
case '%':
fputc('%', file);
length++;
break;
/* %c: print out a character */
case 'c':
char_temp = va_arg(arg, int);
fputc(char_temp, file);
length++;
break;
/* %s: print out a string */
case 's':
string_temp = va_arg(arg, char *);
fputs(string_temp, file);
length += strlen(string_temp);
break;
/* %d: print out an int */
case 'd':
int_temp = va_arg(arg, int);
itoa(int_temp, buffer, 10);
fputs(buffer, file);
length += strlen(buffer);
break;
/* %x: print out an int in hex */
case 'x':
int_temp = va_arg(arg, int);
itoa(int_temp, buffer, 16);
fputs(buffer, file);
length += strlen(buffer);
break;
case 'f':
double_temp = va_arg(arg, double);
ftoa_fixed(buffer, double_temp);
fputs(buffer, file);
length += strlen(buffer);
break;
case 'e':
double_temp = va_arg(arg, double);
ftoa_sci(buffer, double_temp);
fputs(buffer, file);
length += strlen(buffer);
break;
}
}
else {
putc(ch, file);
length++;
}
}
return length;
}
int normalize(double *val) {
int exponent = 0;
double value = *val;
while (value >= 1.0) {
value /= 10.0;
++exponent;
}
while (value < 0.1) {
value *= 10.0;
--exponent;
}
*val = value;
return exponent;
}
static void ftoa_fixed(char *buffer, double value) {
/* carry out a fixed conversion of a double value to a string, with a precision of 5 decimal digits.
* Values with absolute values less than 0.000001 are rounded to 0.0
* Note: this blindly assumes that the buffer will be large enough to hold the largest possible result.
* The largest value we expect is an IEEE 754 double precision real, with maximum magnitude of approximately
* e+308. The C standard requires an implementation to allow a single conversion to produce up to 512
* characters, so that's what we really expect as the buffer size.
*/
int exponent = 0;
int places = 0;
static const int width = 4;
if (value == 0.0) {
buffer[0] = '0';
buffer[1] = '\0';
return;
}
if (value < 0.0) {
*buffer++ = '-';
value = -value;
}
exponent = normalize(&value);
while (exponent > 0) {
int digit = value * 10;
*buffer++ = digit + '0';
value = value * 10 - digit;
++places;
--exponent;
}
if (places == 0)
*buffer++ = '0';
*buffer++ = '.';
while (exponent < 0 && places < width) {
*buffer++ = '0';
--exponent;
++places;
}
while (places < width) {
int digit = value * 10.0;
*buffer++ = digit + '0';
value = value * 10.0 - digit;
++places;
}
*buffer = '\0';
}
void ftoa_sci(char *buffer, double value) {
int exponent = 0;
int places = 0;
static const int width = 4;
if (value == 0.0) {
buffer[0] = '0';
buffer[1] = '\0';
return;
}
if (value < 0.0) {
*buffer++ = '-';
value = -value;
}
exponent = normalize(&value);
int digit = value * 10.0;
*buffer++ = digit + '0';
value = value * 10.0 - digit;
--exponent;
*buffer++ = '.';
for (int i = 0; i < width; i++) {
int digit = value * 10.0;
*buffer++ = digit + '0';
value = value * 10.0 - digit;
}
*buffer++ = 'e';
itoa(exponent, buffer, 10);
}
int my_printf(char const *fmt, ...) {
va_list arg;
int length;
va_start(arg, fmt);
length = my_vfprintf(stdout, fmt, arg);
va_end(arg);
return length;
}
int my_fprintf(FILE *file, char const *fmt, ...) {
va_list arg;
int length;
va_start(arg, fmt);
length = my_vfprintf(file, fmt, arg);
va_end(arg);
return length;
}
#ifdef TEST
int main() {
float floats[] = { 0.0, 1.234e-10, 1.234e+10, -1.234e-10, -1.234e-10 };
my_printf("%s, %d, %x\n", "Some string", 1, 0x1234);
for (int i = 0; i < sizeof(floats) / sizeof(floats[0]); i++)
my_printf("%f, %e\n", floats[i], floats[i]);
return 0;
}
#endif
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