这个程序只可以做1位整数的4则运算和括号内计算，怎么改动可以计算多位整数呢？

#include

#include

int opnd_top=-1,optr_top=-1;

double

这个程序只可以做1位整数的4则运算和括号内计算，怎么改动可以计算多位整数呢？

#include

#include

int opnd_top=-1,optr_top=-1;

double opnd_stack[40];

char optr_stack[40];

char optr[7]={'+', '-' ,'*' ,'/' ,'(' ,')' ,'#'};

char priority[7][7]=

{

{'>','>','','>'},

{'>','>','','>'},

{'>','>','>','>','','>'},

{'>','>','>','>','','>'},

{'

{'>','>','>','>',' ','>','>'},

{'

};

double pop_nd()

{

double e;

e=opnd_stack[opnd_top--];

return e;

}

char pop_tr()

{

char e;

e=optr_stack[optr_top--];

return e;

}

void push_nd(double e)

{

opnd_stack[++opnd_top]=e;

}

void push_tr(char e)

{

optr_stack[++optr_top]=e;

}

double gettop_nd()

{

return opnd_stack[opnd_top];

}

char gettop_tr()

{

return optr_stack[optr_top];

}

int getIndex(char op)

{

int i;

for(i=0;i<7;i++)

if(optr[i]==op) return i;

return -1;

}

char getprior(char optr1,char optr2)

{

int i,j;

i=getIndex(optr1);

j=getIndex(optr2);

if(i==-1 || j==-1)

return ' ';

return priority[i][j];

}

double calculate(double a,double b ,char c)

{

double r=0;

switch©

{

case '+':

r=a+b;

break;

case '-':

r=a-b;

break;

case '*':

r=a*b;

break;

case '/':

r=a/b;

break;

}

return r;

}

void main()

{

char c;

push_tr('#');

printf("input a expression end with #\n");

c=getchar();

while(c!='#' || gettop_tr()!='#')

{

if(c>='0' && c<='9')

{

push_nd((int)(c-'0'));

c=getchar();

}

else

{

switch(getprior(gettop_tr(),c))

{

case '

push_tr©;

c=getchar();

break;

case '=':

pop_tr();

c=getchar();

break;

case '>':

push_nd( calculate(pop_nd(),pop_nd(),pop_tr()) );

break;

}

}

}

printf("the result is: %f\n",pop_nd());

}

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