如果“&"会发生什么不是放在"scanf"语句中? [英] What will happen if '&' is not put in a 'scanf' statement?

问题描述

我去接受采访时被问到以下问题:

I had gone to an interview in which I was asked the question:

您如何看待以下内容?

What do you think about the following?

int i;
scanf ("%d", i);
printf ("i: %d\n", i);

我回答:

• 程序将成功编译.
• 它将打印错误的数字，但是它将一直运行到最后不会崩溃

我的回答是错误的.我不知所措.

The response that I made was wrong. I was overwhelmed.

此后，他们解雇了我

在某些情况下，该程序可能会崩溃并导致核心转储.

The program would crash in some cases and lead to an core dump.

我不明白为什么 程序崩溃?谁能解释我的原因?任何帮助表示赞赏.

I could not understand why the program would crash? Could anyone explain me the reason? Any help appreciated.

推荐答案

定义变量后，编译器会为该变量分配内存.

When a variable is defined, the compiler allocates memory for that variable.

int i;  // The compiler will allocate sizeof(int) bytes for i

上面定义的

i 尚未初始化，并且具有不确定的值.

i defined above is not initialized and have indeterminate value.

要将数据写入分配给 i 的内存位置，您需要指定变量的地址.声明

To write data to that memory location allocated for i, you need to specify the address of the variable. The statement

scanf("%d", &i);

将由用户将 int 数据写入分配给 i 的内存位置.

will write an int data by the user to the memory location allocated for i.

如果未将& 放在 i 之前，则 scanf 将尝试将输入数据写入存储位置 i而不是& i .由于 i 包含不​​确定的值，因此可能存在与内存地址的值相等的值，或者可能包含超出内存地址范围的值.

If & is not placed before i, then scanf will try to write the input data to the memory location i instead of &i. Since i contains indeterminate value, there are some possibilities that it may contain a value equivalent to the value of a memory address or it may contain a value which is out of range of memory address.

在任何一种情况下，程序都可能会出现异常行为，并导致未定义的行为.在这种情况下，什么都可能发生.

In either case, the program may behave erratically and will lead to undefined behavior. In that case anything could happen.

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