The Loop Operation Let's spend a few moments writing a plan as a pseudocode. sum <- 0 number - 0 Grab the current character -> ch If ch is a digit then ● digit <-ascii-to-decimal(ch) number<- number * 10 number<- number + digit Else sum <- sum + number number <- 0 Here are some notes on implementing this plan in C++. Use str.at () to grab the current character; store it in a variable Use isdigit() to see if the character is between '0' and '9' inclusive There is no ascii-to-decimal function. Instead, subtract the character '0' from ch and store the result in the variable digit. Note that the ch has an underlying ASCII code and the codes are sequential. If ch is '0' then subtracting '0' will store the binary number in digit. If the character is '1', then subtracting '0' will leave the binary number 1. Multiply the current value of number by 10, and then add the digit. For instance, if number has the value 2 and digit has the value 5, then number * 10 -> 20 and adding 5 leaves number with 25, which is correct. When you encounter something that isn't a digit, then add the current value of number to sum. Then, set number back to 0 for the next iteration.

The Loop Operation Let's spend a few moments writing a plan as a pseudocode. sum <- 0 number - 0 Grab the current character -> ch If ch is a digit then ● digit <-ascii-to-decimal(ch) number<- number * 10 number<- number + digit Else sum <- sum + number number <- 0 Here are some notes on implementing this plan in C++. Use str.at () to grab the current character; store it in a variable Use isdigit() to see if the character is between '0' and '9' inclusive There is no ascii-to-decimal function. Instead, subtract the character '0' from ch and store the result in the variable digit. Note that the ch has an underlying ASCII code and the codes are sequential. If ch is '0' then subtracting '0' will store the binary number in digit. If the character is '1', then subtracting '0' will leave the binary number 1. Multiply the current value of number by 10, and then add the digit. For instance, if number has the value 2 and digit has the value 5, then number * 10 -> 20 and adding 5 leaves number with 25, which is correct. When you encounter something that isn't a digit, then add the current value of number to sum. Then, set number back to 0 for the next iteration.

C++ for Engineers and Scientists

4th Edition

ISBN:9781133187844

Author:Bronson, Gary J.

Publisher:Bronson, Gary J.

Chapter5: Repetition Statements

Section: Chapter Questions

Problem 9PP

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