我可以使用指针作为C语言中的gets参数吗? [英] Can I use a pointer as the parameter of gets in C?

问题描述

我正在学习C语言.

当我将指针传递给gets()时，我发现它可以很好地工作.

When I pass a pointer to gets(), I find it can work well.

char *ptr;
gets(ptr);
puts(ptr);

但是，如果我定义了一个指针数组，那是行不通的.

But if I define an array of pointers, it doesn't work.

char *ptr[4];
int i=0;
for(i=0;i<4;++i)
    gets(ptr[i]);
for(i=0;i<4;++i)
    puts(ptr[i]);

它们是不同的，还是第一部分实际上是错误的?我想知道原因.

Are they different, or is the first part wrong in fact? I want to know the reason.

推荐答案

您将指针传递给函数(例如gets())，该函数将数据写入指针所指向的存储位置.在您的情况下，您具有 uninitialized 指针，这意味着它指向随机的内存位置(应用程序或操作系统所在的位置).这会导致随机效应-从工作"到异常终止或挂起.您需要保留内存并分配指向该内存的指针，例如通过:

You pass a pointer to a function (e.g. gets()) that writes data to a memory location pointed by your pointer. In your case, you have the uninitialized pointer, which means it points to a random memory location (where applications or an operating system resides). This leads to random effects - from "working" to abnormal termination or hanging. You need to reserve memory and assign pointer to point there, e.g. by:

char *ptr = (char*)malloc(256);

gets(ptr);
puts(ptr);

free(ptr);

考虑使用更安全的gets_s().

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