printf()等系統庫函數是如何實現的？

int printf(FILE *stream, const char *format, va_list arg)
{
int i; //聲明一個變數，-_-
//下面申請一個空間，用來保存將帶有%d，%s之類展開之後的字元串
//例子：("Hello %d 0x%x
", 1 ,10) --&> "Hello 1 0x0a
"
char *s = malloc(4096);
//對於例子里的轉換：
//s是轉換之後字元串的指針
//format指向的是"Hello %d 0x%x
"
//arg是參數列表：1，10（就是從下標為1開始到最後的參數）
i = vsprintf(s, format, arg);
i = fwrite(s, 1, i, stream); //這是個系統調用，這個再去學習，這裡假設成將字元
//串拷貝到數組stream中
free(s); //釋放空間
return i;
}


下面是vsprintf的一個簡單實現

int vsprintf(char *s, const char *format, va_list arg)
{
UCHAR c, *t = s, *p, flag_left, flag_zero /* , flag_sign, flag_space */;
UCHAR temp[32], *q;
temp[31] = "";
int field_min, field_max, i;
long l;
static char hextable_X[16] = "0123456789ABCDEF";
static char hextable_x[16] = "0123456789abcdef";
for (;;) {
c = *format++;
if (c != "%") {
put1char:
*t++ = c;
if (c)
continue;
return t - (UCHAR *) s - 1;
}
flag_left = flag_zero = /* flag_sign = flag_space = flag_another = */ 0;
c = *format++;
for (;;) {
if (c == "-")
flag_left = 1;
else if (c == "0")
flag_zero = 1;
else
break;
c = *format++;
}
field_min = 0;
if ("1" &<= c c &<= "9") { format--; field_min = (int) strtoul0(format, 10, c); c = *format++; } else if (c == "*") { field_min = va_arg(arg, int); c = *format++; } field_max = INT_MAX; if (c == ".") { c = *format++; if ("1" &<= c c &<= "9") { format--; field_min = (int) strtoul0(format, 10, c); c = *format++; } else if (c == "*") { field_max = va_arg(arg, int); c = *format++; } } if (c == "s") { if (field_max != INT_MAX) goto mikan; p = va_arg(arg, char *); l = strlen(p); if (*p) { c = " "; copy_p2t: if (flag_left == 0) { while (l &< field_min) { *t++ = c; field_min--; } } do { *t++ = *p++; } while (*p); } while (l &< field_min) { *t++ = " "; field_min--; } continue; } if (c == "l") { c = *format++; if (c != "d" c != "x" c != "u") { format--; goto mikan; } } if (c == "u") { l = va_arg(arg, UINT); goto printf_u; } if (c == "d") { printf_d: l = va_arg(arg, long); if (l &< 0) { *t++ = "-"; l = - l; field_min--; } printf_u: if (field_max != INT_MAX) goto mikan; if (field_min &<= 0) field_min = 1; p = setdec(temp[31], l); printf_x2: c = " "; l = temp[31] - p; if (flag_zero) c = "0"; goto copy_p2t; } if (c == "i") goto printf_d; if (c == "%") goto put1char; if (c == "x") { q = hextable_x; printf_x: l = va_arg(arg, UINT); p = temp[31]; do { *--p = q[l 0x0f]; } while ((*(UINT *) l) &>&>= 4);
goto printf_x2;
}
if (c == "X") {
q = hextable_X;
goto printf_x;
}
if (c == "p") {
i = (int) va_arg(arg, void *);
p = temp[31];
for (l = 0; l &< 8; l++) { *--p = hextable_X[i 0x0f]; i &>&>= 4;
}
goto copy_p2t;
}
if (c == "o") {
l = va_arg(arg, UINT);
p = temp[31];
do {
*--p = hextable_x[l 0x07];
} while ((*(UINT *) l) &>&>= 3);
goto printf_x2;
}
if (c == "f") {
if (field_max == INT_MAX)
field_max = 6;
/* for ese */
if (field_min &< field_max + 2) field_min = field_max + 2; do { *t++ = "#"; } while (--field_min); continue; } if(c == "c"){ if (field_max == INT_MAX) field_max = 6; /* for ese */ if (field_min &< field_max + 2) field_min = field_max + 2; do { *t++ = "C"; } while (--field_min); continue; } mikan: for(;;); } }

這大概就是printf實現的基本原理，如果你想運行，添加下面的代碼

typedef unsigned char UCHAR;
typedef unsigned int UINT;

static int prefix(int c)
{
signed char base = 0;
if ("a" &<= c c &<= "z") c += "A" - "a"; if (c == "B") base = 2; if (c == "D") base = 10; if (c == "O") base = 8; if (c == "X") base = 16; return base; } unsigned long strtoul0(const unsigned char **ps, int base, unsigned char *errflag) { const unsigned char *s = *ps; unsigned long val = 0, max; int digit; if (base == 0) { base += 10; if (*s == "0") { base = prefix(*(s + 1)); if (base == 0) base += 8; /* base = 8; */ } } if (*s == "0") { if (base == prefix(*(s + 1))) s += 2; } max = ULONG_MAX / base; *errflag = 0; for (;;) { digit = 99; if ("0" &<= *s *s &<= "9") digit = *s - "0"; if ("A" &<= *s *s &<= "Z") digit = *s - ("A" - 10); if ("a" &<= *s *s &<= "z") digit = *s - ("a" - 10); if (digit &>= base)
break;
if (val &> max)
goto err;
val *= base;
if (ULONG_MAX - val &< (unsigned long) digit) { err: *errflag = 1; val = ULONG_MAX; } else val += digit; s++; } *ps = s; return val; } static UCHAR *setdec(UCHAR *s, UINT ui) { do { *--s = (ui % 10) + "0"; } while (ui /= 10); return s; }

int write(int fd, const void *buf, int count)

{

int res = 0;

asm(
"movl $4, %%eax
"

"movl %1, %%ebx
"

"movl %2, %%ecx
"

"movl %3, %%edx
"

"int $0x80
"

"movl %%eax, %0
"

:"=m"(res)

:"m"(fd),

"m"(buf), "m"(count)
);

return res;

}

(知乎的編輯器略噁心）

從一個printf()函數足以窺探一下整個計算機體系的秘密。

首先推薦南七技校網路課 Linux內核分析 - 網易雲課堂 這門課講的很好，擼下來大有幫助

如果題主很貧窮（這門課內部價也不低），可以在網上搜羅這門課的上課筆記。

題主的這個問題可以在系統調用過程 - 20135124 - 博客園 領會。

基本上串聯起來就是 高級語言——系統API(運行庫)——中斷——系統調用——硬體資源

為什麼說此問題足以窺探整個計算機體系的秘密。

說到系統調用得說內核態吧，說到切換得說到中斷吧，說到中斷得說到向量表吧，說到向量表得說到BIOS吧(Linux中系統跟bios的關係其實不是特別大)，說到BIOS得說到ROM吧，說到ROM得說到觸發器吧，說到觸發器得說到邏輯門吧，說到邏輯門得說到晶體管吧，說到晶體管得說到不純半導體吧。。。。。。硬體方面，南北橋結構、匯流排、埠定址、訪問控制方式。。。。。。更不用說高級語言層面的東西了。

即能由小知大、又可以由大見小。