的malloc和scanf [英] Malloc and scanf

问题描述

我在一些脚本语言还算称职，但我终于强迫自己去学习生C.我只是用一些基本的东西（I / O现在）玩耍。我该如何分配堆内存，存储在内存中分配一个字符串，然后吐出来背出来了呢？这就是我现在所拥有的，我怎样才能使它正常工作？

 的#include＆LT;＆stdio.h中GT;
＃包括LT＆;＆stdlib.h中GT;INT主（INT ARGC，CHAR *的argv []）
{
  字符* toParseStr =（字符*）malloc的（10）;
  scanf函数（输入一个字符串，＆安培; toParseStr）;
  的printf（％S，toParseStr）;
  返回0;
}
 

目前我越来越喜欢怪异的输出'8'\\'。

解决方案

 的char * toParseStr =（字符*）malloc的（10）;
  的printf（请在此处输入字符串：）;
  scanf函数（％S，toParseStr）;
  的printf（％S，toParseStr）;
  免费（toParseStr）;
 

首先，在 scanf函数的字符串指定它要接收输入。为了接收键盘输入之前显示的字符串，用的printf ，如图所示。

其次，你不需要解引用 toParseStr ，因为它是你与的malloc 。 如果你使用的是功能，将它指向另一个内存位置，，然后 ＆放大器; toParseStr 要求

例如，假设你想写一个函数来分配内存。然后，你需要＆放大器; toParseStr ，因为你改变指针变量的内容（这是内存中的地址 - 您可以通过打印看到自己的内容）。

 无效AllocateString（字符** ptr_string，const int的N）
{
    * ptr_string =（字符*）malloc的（N）
}
 

正如你所看到的，它接受的char ** ptr_string 其内容，它储存的指针的内存位置的指针会商店现在的 N 字节的分配块的第一个字节（内存地址（的malloc后操作）它自从它的一些垃圾内存地址时不被初始化）。

  INT主（INT ARGC，CHAR *的argv []）
{
  字符* toParseStr;
  const int的N = 10;
  的printf（垃圾：％P \\ N，toParseStr）;
  AllocateString（安培; toParseStr，N）;
  的printf（％d字节一个连续数组的第一个元素的地址：％P \\ N，N，toParseStr）;  的printf（请在此处输入字符串：）;
  scanf函数（％S，toParseStr）;
  的printf（％S，toParseStr）;
  免费（toParseStr）;  返回0;
}
 

第三，建议您分配的可用内存。虽然这是你的整个程序，当程序退出该内存将被释放，它仍然是很好的做法。

I'm fairly competent in a few scripting languages, but I'm finally forcing myself to learn raw C. I'm just playing around with some basic stuff (I/O right now). How can I allocate heap memory, store a string in the allocated memory, and then spit it back out out? This is what I have right now, how can I make it work correctly?

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
  char *toParseStr = (char*)malloc(10);
  scanf("Enter a string",&toParseStr);
  printf("%s",toParseStr);
  return 0;
}

Currently I'm getting weird output like '8'\'.

解决方案

  char *toParseStr = (char*)malloc(10);
  printf("Enter string here: ");
  scanf("%s",toParseStr);
  printf("%s",toParseStr);
  free(toParseStr);

Firstly, the string in scanf is specifies the input it's going to receive. In order to display a string before accepting keyboard input, use printf as shown.

Secondly, you don't need to dereference toParseStr since it's pointing to a character array of size 10 as you allocated with malloc. If you were using a function which would point it to another memory location, then &toParseStr is required.

For example, suppose you wanted to write a function to allocate memory. Then you'd need &toParseStr since you're changing the contents of the pointer variable (which is an address in memory --- you can see for yourself by printing its contents).

void AllocateString(char ** ptr_string, const int n)
{
    *ptr_string = (char*)malloc(n)
}

As you can see, it accepts char ** ptr_string which reads as a pointer which stores the memory location of a pointer which will store the memory address (after the malloc operation) of the first byte of an allocated block of n bytes (right now it has some garbage memory address since it is uninitialized).

int main(int argc, char *argv[])
{
  char *toParseStr;
  const int n = 10;
  printf("Garbage: %p\n",toParseStr);
  AllocateString(&toParseStr,n);
  printf("Address of the first element of a contiguous array of %d bytes: %p\n",n,toParseStr);

  printf("Enter string here: ");
  scanf("%s",toParseStr);
  printf("%s",toParseStr);
  free(toParseStr);

  return 0;
}

Thirdly, it is recommended to free memory you allocate. Even though this is your whole program, and this memory will be deallocated when the program quits, it's still good practice.

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