sprintf或printf的最小实现
我正在研究一种速度至关重要的嵌入式DSP，而且内存很短。
目前，sprintf使用我代码中任何函数的大部分资源。 我只用它来格式化一些简单的文本： %d, %e, %f, %s ，没有精确或奇异的操作。
如何实现更适合我的使用的基本sprintf或printf函数？
这个假定存在一个itoa将int转换为字符表示，并且fputs将字符串写出到你希望它去的地方。
浮点输出至少在一个方面是不符合的：它没有像标准要求那样正确地进行舍入，所以如果你有（例如）内部存储为1.2399999774的值1.234 ，它将会打印为1.2399而不是1.2340 。 这节省了相当多的工作量，并且对于大多数典型用途而言仍然足够。
除了你询问的转换之外，这还支持%c和%x ，但是如果你想要删除它们，那么删除它们是非常简单的（这样做显然可以节省一些内存）。
#include #include #include #include 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