以下 C 代码的运行结果是什么() #include <stdio.h> int main() { int numbers[5]; int* p; p = numbers; *p = 10; p++; *p = 20; p = &numbers[2]; *p = 30; p = numbers + 3; *p = 40; p = numbers; *(p + 4) = 50; for (int n = 0; n < 5; n++) printf(“%d”, numbers[n]); printf(” “); return 0; }
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以下 C 代码的运行结果是什么()
#include <stdio.h> int main() { int numbers[5]; int* p; p = numbers; *p = 10; p++; *p = 20; p = &numbers[2]; *p = 30; p = numbers + 3; *p = 40; p = numbers; *(p + 4) = 50; for (int n = 0; n < 5; n++) printf("%d", numbers[n]); printf(" "); return 0; }
没有括号是重点,没有括号,循环体就是下一行
37:04
以上就是关于问题以下 C 代码的运行结果是什么() #include <stdio.h> int main() { int numbers[5]; int* p; p = numbers; *p = 10; p++; *p = 20; p = &numbers[2]; *p = 30; p = numbers + 3; *p = 40; p = numbers; *(p + 4) = 50; for (int n = 0; n < 5; n++) printf(“%d”, numbers[n]); printf(” “); return 0; }的答案
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qklbishe.com区块链毕设代做网专注|以太坊fabric-计算机|java|毕业设计|代做平台-javagopython毕设 » 以下 C 代码的运行结果是什么() #include <stdio.h> int main() { int numbers[5]; int* p; p = numbers; *p = 10; p++; *p = 20; p = &numbers[2]; *p = 30; p = numbers + 3; *p = 40; p = numbers; *(p + 4) = 50; for (int n = 0; n < 5; n++) printf(“%d”, numbers[n]); printf(” “); return 0; }
微信:btc9767
TELEGRAM :https://t.me/btcok9
具体资料介绍
web3的一级市场千万收益的逻辑
进群点我
qklbishe.com区块链毕设代做网专注|以太坊fabric-计算机|java|毕业设计|代做平台-javagopython毕设 » 以下 C 代码的运行结果是什么() #include <stdio.h> int main() { int numbers[5]; int* p; p = numbers; *p = 10; p++; *p = 20; p = &numbers[2]; *p = 30; p = numbers + 3; *p = 40; p = numbers; *(p + 4) = 50; for (int n = 0; n < 5; n++) printf(“%d”, numbers[n]); printf(” “); return 0; }
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qklbishe.com区块链毕设代做网专注|以太坊fabric-计算机|java|毕业设计|代做平台-javagopython毕设 » 以下 C 代码的运行结果是什么() #include <stdio.h> int main() { int numbers[5]; int* p; p = numbers; *p = 10; p++; *p = 20; p = &numbers[2]; *p = 30; p = numbers + 3; *p = 40; p = numbers; *(p + 4) = 50; for (int n = 0; n < 5; n++) printf(“%d”, numbers[n]); printf(” “); return 0; }
qklbishe.com区块链毕设代做网专注|以太坊fabric-计算机|java|毕业设计|代做平台-javagopython毕设 » 以下 C 代码的运行结果是什么() #include <stdio.h> int main() { int numbers[5]; int* p; p = numbers; *p = 10; p++; *p = 20; p = &numbers[2]; *p = 30; p = numbers + 3; *p = 40; p = numbers; *(p + 4) = 50; for (int n = 0; n < 5; n++) printf(“%d”, numbers[n]); printf(” “); return 0; }