C语言 sprintf 或 printf 的最小实现
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Minimal implementation of sprintf or printf
提问by Gui13
I'm working on an embedded DSP where speed is crucial, and memory is very short.
我正在研究速度至关重要且内存非常短的嵌入式 DSP。
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.
目前，在我的代码中，sprintf 使用的资源最多。我只用它来格式化一些简单的文本：%d, %e, %f, %s，没有任何精确或奇异的操作。
How can I implement a basic sprintf or printf function that would be more suitable for my usage?
如何实现更适合我的使用的基本 sprintf 或 printf 函数？
回答by Jerry Coffin
This one assumes the existence of an itoato convert an int to character representation, and an fputsto write out a string to wherever you want it to go.
这个假设存在itoa将 int 转换为字符表示，以及将fputs字符串写出到任何你想要它去的地方。
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 1.234that is internally stored as 1.2399999774, it'll be printed out as 1.2399instead of 1.2340. This saves quite a bit of work, and remains sufficient for most typical purposes.
浮点输出至少在一个方面不符合标准：它不会按照标准要求正确舍入，因此如果您有（例如）1.234内部存储为的值1.2399999774，它将被打印out as1.2399而不是1.2340. 这节省了相当多的工作，并且对于大多数典型用途仍然足够。
This also supports %cand %xin 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 littlememory).
这也支持%c并且%x除了您询问的转换之外，但是如果您想摆脱它们，删除它们非常简单（这样做显然会节省一点内存）。
#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] = '##代码##';
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 = '##代码##';
}
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] = '##代码##';
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
