06 2021 档案
摘要：#include <stdio.h> #include <string.h> #define MSG "woshishuya" int main() { char arr_str[] = MSG; char * pt_str = MSG; printf("1.数组的pointer:%p\n", ar
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摘要：#include <stdio.h> void print_arr(int size,int arr[size]); int main() { int i_arr[10] = { 1, 2, 3, 4, 5, 6, 6, 8, 9, 10 }; print_arr(10,i_arr); //匿名数组
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摘要：#include <stdio.h> void print_arr(int size,int arr[size]); int main() { int i_arr[10] = { 1, 2, 3, 4, 5, 6, 6, 8, 9, 10 }; print_arr(10,i_arr); return
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摘要：#include <stdio.h> void print_arr(int (* pt)[2] ,int size); void print_pointer(int (* pt)[2] ,int size); int main() { int t_arr[4][2] = { { 1, 200 },
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摘要：#include <stdio.h> int main() { int t_arr[4][2] = { { 1, 2 }, { 3, 4 }, { 5, 6 }, { 7,8 } }; int o_arr[4] = { 1, 2, 3, 4 }; //获取元素值 printf("数组表示法-[3][
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摘要：#include <stdio.h> void getsum(int pt[],int size ); void getsum1(int * head,int * tail ); int main() { int i_arr[5] = { 1, 20, 3, 4, 5 }; int *pt; get
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摘要：#include <stdio.h> unsigned int Loop_Fibonacci(unsigned int n); unsigned int Recu_Fibonacci(unsigned int n); int main() { //给定正整数n(项数),返回响应的斐波那契数值 //斐
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摘要：#include <stdio.h> int to_binary(int a); int main() { //求一个整数(十进制)的二进制(循环语句版) int a = 1234; int result; int shang; int size; int re; int b = a; int c
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摘要：#include <stdio.h> long fact(int a); //计算指定参数的阶乘 int main() { //用循环求a的阶乘(注意输入参数不要超过12,会超过long的范围) int a = 4; long resutl = 1; for (int i = 1; i <= a;
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