# Order of operations

## Order of operations

is read as "Divide 10 BY 5". Thus the solution is 2, not ½ as would be the result of an incorrect analysis.

Prefix notation is especially popular with stack-based operations due to its innate ability to easily distinguish order of operations without the need for parentheses. To evaluate order of operations under prefix notation, one does not even need to memorize an operational hierarchy, as with infix notation. Instead, one looks directly to the notation to discover which operator to evaluate first. Reading an expression from left to right, one first looks for an operator and proceeds to look for two operands. If another operator is found before two operands are found, then the old operator is placed aside until this new operator is resolved. This process iterates until an operator is resolved, which must happen eventually, as there must be one more operand than there are operators in a complete statement. Once resolved, the operator and the two operands are replaced with a new operand. Because one operator and two operands are removed and one operand is added, there is a net loss of one operator and one operand, which still leaves an expression with N operators and N+1 operands, thus allowing the iterative process to c

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Prefix notation is especially popular with stack-based operations due to its innate ability to easily distinguish order of operations without the need for parentheses. To evaluate order of operations under prefix notation, one does not even need to memorize an operational hierarchy, as with infix notation. Instead, one looks directly to the notation to discover which operator to evaluate first. Reading an expression from left to right, one first looks for an operator and proceeds to look for two operands. If another operator is found before two operands are found, then the old operator is placed aside until this new operator is resolved. This process iterates until an operator is resolved, which must happen eventually, as there must be one more operand than there are operators in a complete statement. Once resolved, the operator and the two operands are replaced with a new operand. Because one operator and two operands are removed and one operand is added, there is a net loss of one operator and one operand, which still leaves an expression with N operators and N+1 operands, thus allowing the iterative process to c

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