Answer:
#include <stdio.h>
void interchangeCase(char phrase[],char c){
for(int i=0;phrase[i]!='\0';i++){
if(phrase[i]==c){
if(phrase[i]>='A' && phrase[i]<='Z')
phrase[i]+=32;
else
phrase[i]-=32;
}
}
}
int main(){
char c1[]="Eevee";
interchangeCase(c1,'e');
printf("%s\n",c1);
char c2[]="Eevee";
interchangeCase(c2,'E');
printf("%s\n",c2);
}
Explanation:
- Create a function called interchangeCase that takes the phrase and c as parameters.
- Run a for loop that runs until the end of phrase and check whether the selected character is found or not using an if statement.
- If the character is upper-case alphabet, change it to lower-case alphabet and otherwise do the vice versa.
- Inside the main function, test the program and display the results.
Answer:
Option d is the correct answer for the above question.
Explanation:
- The first loop of the program has a second loop and then the statement. In this scenario, the second loop executes for the value of the first loop and the statement executes for the value of the second loop.
- The first loop executes 4 times, Then the second loop or inner loop executes n times for the n iteration of the first loop, for example, 1 time for the first iteration of the first loop, 2 times for the second iteration of the first loop and so on.
- Then the inner loop executes (1+2+3+4) iteration which gives the result 10 iterations.
- The sum initial value is 0 and the "sum++", increase the value of the sum by 1.
- So the value of the sum becomes 10 after completing 10 iterations of the inner for loop.
- Hence the 10 will be the output. So the Option d is the correct answer while the other is not.
Answer:
July 9, 2015
Explanation:
It was released on macOS on July 9, 2015, and on Microsoft Windows on September 22, 2015, for Office 365 subscribers. Mainstream support ended on October 13, 2020, and extended support for most editions of Office 2016 will end on October 14, 2025, the same day as Windows 10.
The combination of two or more networks is internet work!!
Hope this helps
The answer to the question is true
