static bool execution_order(const TCHAR *input, double *outval)
{
const TCHAR *strpos = input, *strend = input + _tcslen(input);
TCHAR c, res[4];
unsigned int sl = 0, rn = 0;
struct calcstack stack[STACK_SIZE] = { {0, .0} }, *sc, *sc2;
double val = 0;
int i;
bool ok = false;
// While there are input tokens left
while(strpos < strend) {
if (sl >= STACK_SIZE)
return false;
// Read the next token from input.
c = *strpos;
// If the token is a value or identifier
if(is_ident(c)) {
// Push it onto the stack.
stack[sl].s = chartostack (c);
++sl;
}
// Otherwise, the token is an operator (operator here includes both operators, and functions).
else if(is_operator(c) || is_function(c)) {
_stprintf(res, _T("_%02d"), rn);
calc_log ((_T("%s = "), res));
++rn;
// It is known a priori that the operator takes n arguments.
unsigned int nargs = op_arg_count(c);
// If there are fewer than n values on the stack
if(sl < nargs) {
// (Error) The user has not input sufficient values in the expression.
return false;
}
// Else, Pop the top n values from the stack.
// Evaluate the operator, with the values as arguments.
if(is_function(c)) {
calc_log ((_T("%c("), c));
while(nargs > 0){
sc = &stack[sl - nargs]; // to remove reverse order of arguments
if(nargs > 1) {
calc_log ((_T("%s, "), sc));
}
else {
calc_log ((_T("%s)\n"), sc));
}
--nargs;
}
sl-=op_arg_count(c);
}
else {
if(nargs == 1) {
sc = &stack[sl - 1];
sl--;
val = docalc1 (c, sc, val);
calc_log ((_T("%c %s = %f;\n"), c, stacktostr(sc), val));
}
else {
sc = &stack[sl - 2];
calc_log ((_T("%s %c "), stacktostr(sc), c));
sc2 = &stack[sl - 1];
val = docalc2 (c, sc, sc2);
sl--;sl--;
calc_log ((_T("%s = %f;\n"), stacktostr(sc2), val));
}
}
// Push the returned results, if any, back onto the stack.
stack[sl].val = val;
stack[sl].s = NULL;
++sl;
}
++strpos;
}
// If there is only one value in the stack
// That value is the result of the calculation.
if(sl == 1) {
sc = &stack[sl - 1];
sl--;
calc_log ((_T("result = %f\n"), val));
if (outval)
*outval = val;
ok = true;
}
for (i = 0; i < STACK_SIZE; i++)
xfree (stack[i].s);
// If there are more values in the stack
// (Error) The user input has too many values.
return ok;
}
bool execution_order(const char *input) {
printf("order: (arguments in reverse order)\n");
const char *strpos = input, *strend = input + strlen(input);
char c, res[4];
unsigned int sl = 0, sc, stack[32], rn = 0;
// While there are input tokens left
while(strpos < strend) {
// Read the next token from input.
c = *strpos;
// If the token is a value or identifier
if(is_ident(c)) {
// Push it onto the stack.
stack[sl] = c;
++sl;
}
// Otherwise, the token is an operator (operator here includes both operators, and functions).
else if(is_operator(c) || is_function(c)) {
sprintf(res, "_%02d", rn);
printf("%s = ", res);
++rn;
// It is known a priori that the operator takes n arguments.
unsigned int nargs = op_arg_count(c);
// If there are fewer than n values on the stack
if(sl < nargs) {
// (Error) The user has not input sufficient values in the expression.
return false;
}
// Else, Pop the top n values from the stack.
// Evaluate the operator, with the values as arguments.
if(is_function(c)) {
printf("%c(", c);
while(nargs > 0) {
sc = stack[sl - 1];
sl--;
if(nargs > 1) {
printf("%s, ", &sc);
}
else {
printf("%s)\n", &sc);
}
--nargs;
}
}
else {
if(nargs == 1) {
sc = stack[sl - 1];
sl--;
printf("%c %s;\n", c, &sc);
}
else {
sc = stack[sl - 1];
sl--;
printf("%s %c ", &sc, c);
sc = stack[sl - 1];
sl--;
printf("%s;\n",&sc);
}
}
// Push the returned results, if any, back onto the stack.
stack[sl] = *(unsigned int*)res;
++sl;
}
++strpos;
}
// If there is only one value in the stack
// That value is the result of the calculation.
if(sl == 1) {
sc = stack[sl - 1];
sl--;
printf("%s is a result\n", &sc);
return true;
}
// If there are more values in the stack
// (Error) The user input has too many values.
return false;
}
