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ss7ja [257]
4 years ago
15

Write a program whose input is a character and a string, and whose output indicates the number of times the character appears in

the string. You may assume that the string does not contain spaces and will always contain less than 50 characters.Ex: If the input is:n Mondaythe output is:1Ex: If the input is:z TodayisMondaythe output is:0Ex: If the input is:n It'ssunnytodaythe output is:2Case matters.Ex: If the input is:n Nobodythe output is:0n is different than N.C++ Code:#include #include int main(void) {/* Type your code here. */return 0;}

Computers and Technology
1 answer:
Zina [86]4 years ago
6 0

Answer:

Here is the C++ program:

#include <iostream> // to include input output functions

using namespace std; // to identify objects like cin cout

int counter(string userString, char character) { //function counter

   int count = 0;   // counts the no of times a character appears in the string

   for (int i=0;i<userString.length();i++)  

// loop to move through the string to find the occurrence of the character

       if (userString[i] == character) //if characters is found in the string

           count++;   //counts the occurrence of the character in the string

   return count; }   //returns the no of times character occurs in the string

int main() { //start of the main() function body

   string s; // stores the string entered by the user

   cout<<"Enter a string: "; //prompts user to enter the string

   cin>>s; //reads the string from user

   char ch; //stores the character entered by the user

   cout<<"Enter a character: "; //prompts user to enter a character

   cin>>ch; //reads the character from user

   cout << counter(s, ch) << endl; }  

//calls counter function to find the number of times a character occurs in the //string

Explanation:

The counter function works as following:

It has a count variable which stores the number of occurrences of a character in the userString.

It uses a for loop which loops through the entire string.

It has i position variable which starts with the first character of the string and checks if the first character of userString matches with the required character.

If it matches the character then count variable counts the first occurrence of the character and in the userString and is incremented to 1.

If the character does not match with the first character of the userString then the loops keeps traversing through the userString until the end of the userString is reached which is specified by the length() function which returns the length of the string.

After the loop ends the return count statement is used to return the number of occurrences of the character in the userString.

The main() function prompts the user to enter a string and a character. It then calls counter() function passing string s and character ch arguments to it in order to get the number of times ch appears in s.

The output is attached in a screenshot.

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Answer:

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Explanation:

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You are tasked to calculate a specific algebraic expansion, i.e. compute the value of f and g for the expression: ???? = (??????
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Answer:

.data

prompt: .asciiz "Enter 4 integers for A, B, C, D respectively:\n"

newLine: .asciiz "\n"

decimal: .asciiz "f_ten = "

binary: .asciiz "f_two = "

decimal2: .asciiz "g_ten = "

binary2: .asciiz "g_two = "

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main:

#display prompt

li $v0, 4

la $a0, prompt

syscall

#Read A input in $v0 and store it in $t0

li $v0, 5

syscall

move $t0, $v0

#Read B input in $v0 and store it in $t1

li $v0, 5

syscall

move $t1, $v0

#Read C input in $v0 and store it in $t2

li $v0, 5

syscall

move $t2, $v0

#Read D input in $v0 and store it in $t3

li $v0, 5

syscall

move $t3, $v0

#Finding A^4

#Loop (AxA)

li $t6, 0

L1:

bge $t6, $t0, quit

add $s1, $s1, $t0 # A=S+A => $s1= A^2

addi $t6, $t6, 1 # i=i+1

j L1

quit:

#Loop (A^2 x A^2)

li $t6, 0

L1A:

bge $t6, $s1, quit1A

add $s5, $s5, $s1

addi $t6,$t6, 1

j L1A

#End of Finding A^4

#Finding 4xA^3

quit1A:

#Loop (4xB)

li $t6, 0

L2:

bge $t6, 4, quit2

add $s2, $s2, $t1

addi $t6, $t6, 1

j L2

quit2:

#Loop (BxB)

li $t6 , 0

L2A:

bge $t6, $t1, quit2A #loop2

add $s6, $s6, $t1 #add

addi $t6, $t6, 1 #add immediate

j L2A #loop2

quit2A: # perform proper program termination using syscall for exit

#Loop (BxB)

li $t6 , 0 #load immediate

L2AA:

bge $t6, $s2, quit2AA #loop2

add $t7, $t7, $s6 #add

addi $t6, $t6, 1 #add immediate

j L2AA #loop2

#End ofFinding 4xA^3

#Finding 3xC^2

quit2AA: # perform proper program termination using syscall for exit

#3 Loop (3 x (C x C)) FOR S3

li $t6 , 0 #load immediate

L3:

bge $t6, $t2, quit3 #loop3

add $s3, $s3, $t2 #add

addi $t6,$t6, 1 #add immediate

j L3 #loop3

quit3: # perform proper program termination using syscall for exit

#3 Loop (3 x (C x C)) FOR S3

li $t6 , 0 #load immediate

L3A:

bge $t6, 3, quit3A #loop3

add $s0, $s0, $s3 #add

addi $t6,$t6, 1 #add immediate

j L3A #loop3

#End of Finding 3xC^2

#Finding 2xD

quit3A: # perform proper program termination using syscall for exit

#4 Loop (2 x D) FOR S4

li $t6 , 0

L4:

bge $t6, 2, quit4 #loop4

add $s4, $s4, $t3 #add

addi $t6, $t6, 1 #add immediate

j L4 #Loop4

#End of Finding 2xD

#Finding AxB^2

quit4:

li $t6, 0

li $s1, 0

L5:

bge $t6, $t1, quit5

add $s1, $s1, $t1

addi $t6, $t6, 1

j L5

quit5:

li $t6, 0

li $s2, 0

L6:

bge $t6, $t0, quit6

add $s2, $s2, $s1

addi $t6, $t6, 1

j L6

#End of Finding AxB^2

#Finding C^2XD^3

quit6: #finds C^2

li $t6, 0

li $s1, 0

L7:

bge $t6, $t2, quit7

add $s1, $s1, $t2

addi $t6, $t6, 1

j L7

quit7: #finds D^2

li $t6, 0

li $s6, 0

L8:

bge $t6, $t3, quit8

add $s6, $s6, $t3

addi $t6, $t6, 1

j L8

quit8: #finds D^3

li $t6, 0

li $s7, 0

L9:

bge $t6, $t3, quit9

add $s7, $s7, $s6

addi $t6, $t6, 1

j L9

quit9: #finds C^2XD^3

li $t6, 0

li $s3, 0

L10:

bge $t6, $s1, end

add $s3, $s3, $s7

addi $t6, $t6, 1

j L10

#End of Finding C^2XD^3

end: # perform proper program termination using syscall for exit

#f is $t8

li $t8 , 0

sub $t8, $s5, $t7 # addition

add $t8, $t8, $s0 # subract

sub $t8,$t8, $s4 # subract

#g is $t9

li $t9 , 0

add $t9, $s2, $s3 # addition

#Display

#1st equation

li $v0,4 # display the answer string with syscall having $v0=4

la $a0, decimal # Gives answer in decimal value

syscall # value entered is returned in register $v0

li $v0, 1 # display the answer string with syscall having $v0=1

move $a0, $t8 # moves the value from $a0 into $t8

syscall # value entered is returned in register $v0

li $v0,4 # display the answer string with syscall having $v0=4

la $a0, newLine # puts newLine in between answers

syscall # value entered is returned in register $v0

li $v0,4 # display the answer string with syscall having $v0=4

la $a0, binary # Gives answer in binary

syscall # value entered is returned in register $v0

li $v0, 35

move $a0, $t8 # moves the value from into $a0 from $t8

syscall # value entered is returned in register $v0

li $v0,4 # display the answer string with syscall having $v0=4

la $a0, newLine # puts newLine in between answers

syscall # value entered is returned in register $v0

#2nd equation

li $v0,4 # display the answer string with syscall having $v0=4

la $a0, decimal2 # Gives answer in decimal value

syscall # value entered is returned in register $v0

li $v0, 1 # display the answer string with syscall having $v0=1

move $a0, $t9 # moves the value from $a0 into $t8

syscall # value entered is returned in reg $v0

li $v0,4 # display the answer string with syscall having $v0=4

la $a0, newLine # puts newLine in between answers

syscall # value entered is returned in register $v0

li $v0,4 # display the answer string with syscall having $v0=4

la $a0, binary2 # Gives answer in binary

syscall # value entered is returned in register $v0

li $v0, 35

move $a0, $t9 # moves the value from into $a0 from $t8

syscall # value entered is returned in register $v0

li $v0,4 # display the answer string with syscall having $v0=4

la $a0, newLine # puts newLine in between answers

syscall # value entered is returned in register $v0

#end the program

li $v0, 10

syscall

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Answer:

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Explanation:

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