本题可变为: 第10章第5节 int max (int x, int y) Rreturn(x>y? x: y int min (int x, int y) freturn(x<y? x y), 3 int add (int x, int y) freturn(x+y) int process( int x,int y, int(*p)o) fint resulte resulte=(p)(x y) return(resulte):) maino fint max(intx, int y), min(int x, int y), add(int x, int y), process( int x, int y int(*p)O); int a b int(*1)0,(j0(k)0; printf("input a, b=); scanf(%d,%d”,&a1&b) -max; j=min; k=add printf(" max=%dn,, process(a, b, i)) printf("max=%dn, process(a, b,D); printf("max=%d\n",process(a, b, k)main() {int max(intx,int y), min(int x, int y), add (int x, int y), process( int x,int y int (*p)()); int a,b int (*i)(), (*j)(), (*k)(); printf(“input a,b=“); scanf(“%d,%d”,&a,&b); i=max; j=min; k=add; printf(“max=%d\n”,process (a,b,i)); printf(“max=%d\n”,process (a,b,j)); printf(“max=%d\n”,process (a,b,k)); } 本题可变为： int max (int x, int y) {return(x>y? x:y);} int min (int x, int y) {return(x<y? x:y);} int add (int x, int y) {return(x+y);} int process( int x,int y , int (*p)( )) {int resulte; resulte=(*p)(x,y); return(resulte); } 第10章第5节