Computers and Technology

What are the key differences between c-style strings and string data type?

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Computers and Technology, 22.06.2019 13:30, ashleypere99
Jane’s team is using the v-shaped model for their project. during the high-level design phase of the project, testers perform integration testing. what is the purpose of an integration test plan in the v-model of development? a. checks if the team has gathered all the requirements b. checks how the product interacts with external systems c. checks the flow of data in internal modules d. checks how the product works from the client side
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Computers and Technology, 22.06.2019 17:30, uh8hardiek
Ou listened to a song on your computer. did you use hardware or software?
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Computers and Technology, 23.06.2019 02:00, mayapril813
Consider the following function main: int main() { int alpha[20]; int beta[20]; int matrix[10][4]; . . } a. write the definition of the function inputarray that prompts the user to input 20 numbers and stores the numbers into alpha. b. write the definition of the function doublearray that initializes the elements of beta to two times the corresponding elements in alpha. make sure that you prevent the function from modifying the elements of alpha. c. write the definition of the function copyalphabeta that stores alpha into the first five rows of matrix and beta into the last five rows of matrix. make sure that you prevent the function from modifying the elements of alpha and beta. d. write the definition of the function printarray that prints any onedimensional array of type int. print 15 elements per line. e. write a c11 program that tests the function main and the functions discussed in parts a through d. (add additional functions, such as printing a two-dimensional array, as needed.)
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Computers and Technology, 24.06.2019 00:40, sierravick123owr441
Use a software program or a graphing utility with matrix capabilities to solve the system of linear equations using an inverse matrix. x1 + 2x2 βˆ’ x3 + 3x4 βˆ’ x5 = 6 x1 βˆ’ 3x2 + x3 + 2x4 βˆ’ x5 = βˆ’6 2x1 + x2 + x3 βˆ’ 3x4 + x5 = 3 x1 βˆ’ x2 + 2x3 + x4 βˆ’ x5 = βˆ’3 2x1 + x2 βˆ’ x3 + 2x4 + x5 = 5
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