Output the decimal character value of the integer variable

The implementation of the C programming language 2nd (ch4.10 recursion) is (parameter n cannot be the smallest integer ):

void printd(int n) {    if (n < 0) {        putchar('-');        n = -n;    }    if (n / 10)        printd(n / 10);    putchar(n % 10 + '0');}

The implementation of C traps and pitfalls (ch7.11 an example of portability problems) is similar:

void printnum(long n, void (*p)(int)) {    if (n < 0) {        (*p)('-');        n = -n;    }    if (n >= 10)        printnum(n / 10, p);    (*p)((int)(n % 10) + '0');}

From the compatibility perspective, Andrew Koenig believes that the last statement has a problem. '0' + 5 get '5'. It is assumed that the computer is implemented based on ANSI, and the characters '0' to '9' are continuous. so it should be changed:

(* P) ("0123456789" [(n % 10)]);

The prerequisite for correct execution of the two programs is that parameter n cannot be the smallest integer. otherwise, n =-N will cause overflow. changing the positive number symbol does not overflow, so Andrew Koenig changes the program to process only negative numbers (if the parameter is a positive number, change its symbol ):

void printneg(long n, void (*p)(int)) {    if (n <= -10)        printneg(n / 10, p);    (*p)("0123456789"[-(n % 10)]);}void printnum(long n, void (*p)(int)) {    if (n < 0) {        (*p)('-');        printneg(n, p);    } else        printneg(-n, p);}

Here is another assumption: the value of N % 10 (n <0) is negative. however, the C language does not define the result symbol. Only the expression r = A % B can be guaranteed. When a> = 0 and B> 0, r> 0 exists, | r | <| B | (ch7.7 how does Division truncate ?). Therefore, you need to modify the printneg function:

void printneg(long n, void (*p)(int)) {    long q;    int r;    q = n / 10;    r = n % 10;    if (r > 0) {        r -= 10;        ++q;    }    if (n <= -10)        printneg(q, p);    (*p)("0123456789"[-r]);}

The following is the execution result (Ubuntu 10.4.1 Linux, 32 bits ):

Printd (0x80000000): -- 214748364 (
Printnum (0x80000000, (void (*) (INT) putchar):-2147483648 references:

The C programming language 2nd (ch4.10 recursion)

C traps and pitfalls (ch7.11 an example of portability problems)