This program takes an arithmetic operator +, -, *, / and two operands from the user. Then, it performs the calculation on the two operands depending upon the operator entered by the user. Simple Calculator using switch Statement #include <stdio.h> int main() { char op; double first, ...
printf("%d is a palindrome.",n); else printf("%d is not a palindrome.",n); return 0; } 结果输出: Enter an integer: 12321 12321 is a palindrome. 3、质数检查 注:1既不是质数也不是合数。 源代码: /* C program to check whether a number is prime or not. */ #include int main()...
for(j=0;j printf("* "); printf("\n"); } return 0; } 5、简单的加减乘除计算器 源代码: /* Source code to create a simple calculator for addition, subtraction, multiplication and division using switch...case statement in C programming. */ # include int main() { char o; float num1...
printf("%d is a palindrome.",n); else printf("%d is not a palindrome.",n); return 0; } 结果输出: Enter an integer: 12321 12321 is a palindrome. 3、质数检查 注:1既不是质数也不是合数。 源代码: /* C program to check whether a number is prime or not. */ #include int main()...
/* Source code to create a simple calculator for addition, subtraction, multiplication and division using switch...case statement in C programming. */ # include <stdio.h> int main() { char o; float num1,num2; printf("Enter operator either + or - or * or divide : "); scanf("%c"...
The program below is used in various topics to demonstrate debugging tasks. This program is a simple calculator that reads its input from a character buffer. If integers are read, they are pushed on a stack. If one of the operators (+ - * ⁄) is read, the top two elements are poppe...
14. A simple calculator program, programming design requirements according to the expression of user input from the keyboard: operand 1 operator op operand 2 calculate the value of the expression, the specified. Operators for addition, subtraction, multiplication, and division Use the time function ...
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/*C program to design love calculator.*/#include <stdio.h>#include <string.h>#include <ctype.h>//function will return sum of all digitsintsumOfDigits(intnum) {intsum=0;while(num>0) { sum+=(num%10); num/=10; }returnsum; }intmain() {charyName[40], pName[40];intsum, sum1...
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