Postfix Evaluation in C
Date : March 29 2020, 07:55 AM
hop of those help? If i try to run your program with 2 + 3, I'm gonna get underflow. The 2 is going to get pushed onto your stack: if(isdigit(eval)){
push((eval-'0'), &infix_stack);
showStack(&infix_stack);
}
opr2 = pop(&infix_stack);
opr1 = pop(&infix_stack);
|
Does type conversion happen if assigning a number to a long type without postfix "L"
Tag : cpp , By : lamberms
Date : March 29 2020, 07:55 AM
wish of those help Yes, it involves conversion. But with a literal like this, the conversion will normally happen at compile time, so appending the L won't make any real difference unless you find the code more readable with the L there (I normally don't).
There are a few cases where you can append a suffix to get a result that's actually different from what you'd get without the suffix, but the ones you've shown don't fall into this category.
|
Postfix evaluation 3
Tag : java , By : Ben Brown
Date : March 29 2020, 07:55 AM
will help you i have worked for this question couple of days and im at the end. Code works well for the input numbers between 0 and 9. but when input is more than 9 for example 28 72 * 13 + 20 67 * + it doesn't print correct answer. I don't know what to implement in order to code work. I cant change main function. I can change only inside of static int evaluatePostfix(char [] izraz, int n) , Instead of Character, try String like below: Replace this code: ArrayStack<Character> e = new ArrayStack(n);
char ch=0,res=0,a,b;
int op1,op2,result=0,c=0;
int z;
for(int i=0; i<n; i++)
{
if(Character.isDigit(izraz[i]))
{
ch=izraz[i];
e.push(ch);
}
if(izraz[i]=='+' || izraz[i]=='-' || izraz[i]=='/' || izraz[i]=='*')
{
ch=izraz[i];
op1 =e.pop()-'0';
//System.out.print(op1+" ");
op2 =e.pop()-'0';
ArrayStack<String> e = new ArrayStack<String>(n);
char ch=0;
int op1,op2,result=0;
for(int i=0; i<n; i++)
{
if(Character.isDigit(izraz[i]))
{
final StringBuilder number= new StringBuilder();
while (izraz[i] != ' ') {
number.append(izraz[i]);
i++;
}
e.push(number.toString());
continue;
}
if(izraz[i]=='+' || izraz[i]=='-' || izraz[i]=='/' || izraz[i]=='*')
{
ch=izraz[i];
op1 =Integer.parseInt(e.pop());
//System.out.print(op1+" ");
op2 =Integer.parseInt(e.pop());
|
I'd like to see one example where the conversion-type-id is looked up in the context of the entire postfix-expression
Date : March 29 2020, 07:55 AM
help you fix your problem In [basic.lookup.classref]/7 (C++14) we have (emphasis is mine): , Something like this: struct C {
operator int() { return 42; }
};
typedef int X;
int main() {
C c;
c.operator X(); // calls c.operator int();
}
|
SQL Server - date time conversion error - The conversion of a varchar data type to a datetime data type resulted in an o
Tag : sql , By : adbanginwar
Date : March 29 2020, 07:55 AM
around this issue That isn't the style 101, it's the style 100 (101 is MM/dd/yyyy). You also have a leading ' ' that you should to remove: SELECT CONVERT(datetime,STUFF(' September 2, 2019 14:29:46',1,1,''),100);
|
Boards Message :
Related QUESTIONS :