C言语作为一种历史长久的编程言语,因其富强的机能跟丰富的库支撑,至今仍被广泛利用于体系开辟、嵌入式体系、游戏开辟等范畴。在东莞,有很多优质的教导资本可能帮助进修者从入门到粗通C言语。本文将具体介绍东莞的基本C言语进修资本、实战剖析以及一些实用的进修技能。
东莞有很多专业培训机构供给C言语课程,以下是一些推荐的机构:
#include <stdio.h>
int main() {
float num1, num2;
char operator;
printf("Enter an operator (+, -, *, /): ");
scanf("%c", &operator);
printf("Enter two operands: ");
scanf("%f %f", &num1, &num2);
switch (operator) {
case '+':
printf("%.1f + %.1f = %.1f", num1, num2, num1 + num2);
break;
case '-':
printf("%.1f - %.1f = %.1f", num1, num2, num1 - num2);
break;
case '*':
printf("%.1f * %.1f = %.1f", num1, num2, num1 * num2);
break;
case '/':
if (num2 != 0.0)
printf("%.1f / %.1f = %.1f", num1, num2, num1 / num2);
else
printf("Division by zero is not allowed.");
break;
default:
printf("Invalid operator!");
}
return 0;
}
#include <stdio.h>
void bubbleSort(int arr[], int n) {
int i, j, temp;
for (i = 0; i < n-1; i++) {
for (j = 0; j < n-i-1; j++) {
if (arr[j] > arr[j+1]) {
temp = arr[j];
arr[j] = arr[j+1];
arr[j+1] = temp;
}
}
}
}
int main() {
int arr[] = {64, 34, 25, 12, 22, 11, 90};
int n = sizeof(arr)/sizeof(arr[0]);
bubbleSort(arr, n);
printf("Sorted array: \n");
for (int i = 0; i < n; i++)
printf("%d ", arr[i]);
printf("\n");
return 0;
}
控制东莞基本C言语须要抉择合适的进修资本,经由过程实战剖析跟技能分享,逐步进步本人的编程才能。只有保持不懈,信赖你必定能从入门到粗通C言语。