c - c 程序的意外行为？

Question

此代码没有按预期提供输出，显然我无法理解程序行为。请帮助我理解这一点。任何帮助将不胜感激。您可以看到程序的输出 https://code.hackerearth.com/31d954z?key=fce7bc8e07fc10a3d7a693e45a5af94a在这里。

1.In the last comment, I can't find why the elements of array are not updated.

2.In the body of func on printing 'a' it gives some unexpected output.

For e.g., if i pass j = 2 and a[0] = 1

       After j = j+1 , which results in j = 3;

       a=a+j should result in a = 4

       but instead it result in a = 13.

#include <stdio.h>

void func(int j,int *a)
{
 j=j+1;
 printf("%d\t%d\n",a,j);     //prints j updated value and a
 a=a+j;
 printf("a = %d\n ",a);      //prints value of a
}


void main()
{
 int a[5] ={1,2,3,4,5},i,j=2;
 for (i =0;i<5;i++ )
 func(j, a[i]);
 for (i =0;i<5;i++ )   
 printf("%d\t",a[i]);    //Prints the array as 1 2 3 4 5
}

运行此代码时，输​​出为:

1 3 // Here a = 1 and j = 3
a = 13 //but after addition a = 13
2 3
a = 14
3 3
a = 15
4 3
a = 16
5 3
a = 17
1 2 3 4 5 //array elements not updated

Answer 1

I want to understand the code behavior.

您的代码产生了未定义的行为，因此您应该停止正在做的事情并调试它。

---

当你想要索引数组时，你可以这样做:

a[i]

i 是索引，a 是你的数组。所以如果你想访问第一个元素，你需要做 a[0]，当你想索引第三个元素时，你做 a[2] 等等上。

---

但是，您可能想要做的只是传递第 i 个元素，添加它并打印它。

因此，您应该启用编译器警告:

prog.c: In function 'func':
prog.c:6:11: warning: format '%d' expects argument of type 'int', but argument 2 has type 'int *' [-Wformat=]
  printf("%d\t%d\n",a,j);     //prints j updated value and a
          ~^
          %ls
prog.c:8:15: warning: format '%d' expects argument of type 'int', but argument 2 has type 'int *' [-Wformat=]
  printf("a = %d\n ",a);      //prints value of a
              ~^
              %ls
prog.c: At top level:
prog.c:12:6: warning: return type of 'main' is not 'int' [-Wmain]
 void main()
      ^~~~
prog.c: In function 'main':
prog.c:16:10: warning: passing argument 2 of 'func' makes pointer from integer without a cast [-Wint-conversion]
  func(j, a[i]);
          ^
prog.c:3:6: note: expected 'int *' but argument is of type 'int'
 void func(int j,int *a)
      ^~~~

然后相应地修改您的代码，例如:

#include <stdio.h>

void func(int j,int a)
{
 j=j+1;
 printf("%d\t%d\n",a,j);   
 a=a+j;
 printf("a = %d\n ",a);
}


int main(void)
{
 int a[5] ={1,2,3,4,5},i,j=2;
 for (i =0;i<5;i++ )
 func(j, a[i]);
 for (i =0;i<5;i++ )   
 printf("%d\t",a[i]);
}

哪些输出:

1   3
a = 4
 2  3
a = 5
 3  3
a = 6
 4  3
a = 7
 5  3
a = 8
 1  2   3   4   5