Logic order of operators

[xra7en]

does multiplication still processed before addition and subtraction?

[Gets]

Code: QB64: [Select]

1.  
2. a(1, 1) = 1
3. b.q1 = 5
4. c = 3
5.  
6. PRINT b.q1 + a(1, 1) * c
7.  
8.  

prints "8", so multiplication comes first.

The code snippet you edited out didn't provide much information, so I can only guess that  some variable isn't being passed to it correctly. Add a printout of the active variables and if statement equations to the error message  to see if you can isolate anything going wrong, I guess?

[xra7en]

after a second look I do not think it is a logic issue as it is where I am placing the operators to check, I need to rearrange the if/then/else
normally I do not have to use parentheses as the logic order works fine.

[freetrav]

Unless documented otherwise, you should always assume - in any language - that the order of evaluation is as follows:

1. Parentheses (grouping)
2. Function Calls
3. Negation (Unary -)
3. Exponentiation (if available, e.g., FORTRAN **)
4. Multiplication and Division (left to right)
5. Addition and Subtraction (left to right)

Note that, for example, APL explicitly documents that evaluation is strictly right-to-left, not hierarchical.

[TempodiBasic]

Hi
IMHO() let us to manage the order of processing the math operation but we must remember the original order of processing operation
http://qb64.org/wiki/Mathematical_Operations#Basic.27s_Order_of_Operations

here just an example

Code: QB64: [Select]

1. DEFINT A-Z
2. a = 10
3. b = 2
4. c = 3
5.  
6. PRINT "a="; a; " b="; b; " c="; c; " d="; d; " e="; e
7. PRINT "a - b * c= "; a - b * c; " (a-b) * c="; (a - b) * c
8. PRINT " a + b * c= "; a + b * c; "  (a + b) * c= "; (a + b) * c
9. PRINT " a + b / c= "; a + b / c; "  (a + b) / c= "; (a + b) / c
10. PRINT " a + b MOD c= "; a + b MOD c; "  (a + b) MOD c= "; (a + b) MOD c
11. PRINT " a MOD b * c= "; a MOD b * c; "  (a MOD b) * c= "; (a MOD b) * c
12. PRINT "a ^ b *c="; a ^ b * c; "(a ^ b) *c="; (a ^ b) * c
13. PRINT "a ^ b *c="; a ^ b * c; "a ^ (b *c)="; a ^ (b * c)
14.  
15.  

I hope it will clear