为什么我要执行此代码的细分故障？

发布于 2025-01-19 14:39:33 字数 346 浏览 4 评论 0原文

我试图找出这段代码的问题：

int main(int argc, char *argv[])
{
    char *s;
    scanf("%s\n", s);

    char *t = malloc(sizeof(*s) + 1);

    strcpy(t, s);
    t[0] = toupper(t[0]);
    printf("s: %s\n", s);
    printf("t: %s\n", t);

    free(t);
    return 0;
}

我试图将输入的第一个字母大写，例如将 hi! 更改为 Hi!。

原文

I'm trying to figure out the issue with this code:

int main(int argc, char *argv[])
{
    char *s;
    scanf("%s\n", s);

    char *t = malloc(sizeof(*s) + 1);

    strcpy(t, s);
    t[0] = toupper(t[0]);
    printf("s: %s\n", s);
    printf("t: %s\n", t);

    free(t);
    return 0;
}

I was trying to upper case the first alphabet of my input to get for example hi! changed into Hi!.

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琉璃繁缕 2025-01-26 14:39:33

您在char*上使用scanf，而无需先分配内存。这可能会导致细分故障。实际上，以下代码：
```
 ＃include＆lt; stdio.h＆gt;
int main（）
{
    char* str;
    scanf（“％s”，str）;
    printf（“％s \ n”，str）;
    返回0;
}
 
```
有时确实会出现分割故障。取而代之的是，如果要避免malloc，则可以使用char数组而不是char*。您可以使用fgets而不是scanf来防止写入界限，也可以使用scanf本身来检查：
```
  char s [100];
if（scanf（“％99S”，s）！= 1）{ / *错误 * /}
 
```
在您的代码中，在％s之后，您在scanf语句中有一个\ n， 't打算这种行为）。您可以在此处查看它的作用：
其次，在为t分配内存时，您使用的是sizeof（*s），它只是size> sizeof（char）定义为1。相反，您应该使用strlen（s）+1，它将分配一个字节，而不是s的长度。另外，还建议检查t == null。
在调用之前在平台上签名。有关此答案的更多信息： https://stackoverflow.com/a/21805970/18639286 （尽管问题是一个C ++问题，答案实际上引用C标准规格）

我已将上述更改编辑为您的代码以获取以下内容：

int main (int argc,char* argv[])
{

    char s[100];
    if(scanf("%99s",s)!=1)
    {
       fprintf(stderr, "Invalid or missing input\n");
       // scanf should have returned 1 if it successfully read 1 input
    }
    char *t = malloc(strlen(s) + 1); // to store the letters + '\0' at the end
    if(t==NULL)
    {
       fprintf(stderr, "Memory allocation failed\n");
       exit(1);
    }

    strcpy(t, s);
    t[0] = toupper((unsigned char)t[0]);
    printf("s: %s\n", s);
    printf("t: %s\n", t);

    free(t);
    return 0;

}

一些额外的积分：

您也可以使用fgets而不是scanf：
fgets（S，100，STDIN）;作为fgets确实绑定了检查。

另外，fgets读取直到线/文件的结尾或缓冲区已满，因此它将通过输入中的空格读取，但是scanf将停止在空间。或正如Ted Lyngmo在评论中指出的那样， scanf带有％s读取一个单词，而fgets读取一行。

但是，即使您打算只读一个单词，也可以使用fgets，然后使用sscanf来读取使用fgets的字符串中的第一个单词。示例：

char buffer[20];
char str[sizeof buffer]=""; //initialize with an empty string
if(fgets(buffer, sizeof buffer, stdin)==NULL)
{
  fprintf(stderr, "Error reading input\n");
  exit(1);
}
if(sscanf(buffer, "%s", str)!=1)
{
  // if str was not read
  fprintf(stderr, "String not read\n");
  exit(1);
}
// DO stuff with str

在这里，您不需要使用％19S，因为我们知道buffer的大小。

You are using scanf on a char* without first allocating memory to it. This may lead to segmentation fault. In fact, the following code :
```
#include<stdio.h>
int main()
{
    char* str;
    scanf("%s", str);
    printf("%s\n", str);
    return 0;
}
```
does give segmentation fault sometimes. Instead you can use a char array instead of a char* if you want to avoid malloc before scanf . You can use fgets instead of scanf to prevent writing out of bounds or you can check with scanf itself like this:
```
char s[100];
if(scanf("%99s",s) != 1) { /* error */ }
```
Also in your code, you have a \n after %s in the scanf statement, that I have removed(assuming you didn't intend this behaviour) . You can see what that does here: Using "\n" in scanf() in C
Secondly, in allocating memory for t, you are using sizeof(*s) which is just sizeof(char) which is defined to be 1. Instead, you should be using strlen(s)+1 which will allocate one more byte than the length of s which should work fine. Also, checking if t==NULL is also recommended.
Before calling toupper the argument must be type-casted into an unsigned char to avoid the risks of undefined behaviour if the char type is signed on the platform. More on this can be found on this answer: https://stackoverflow.com/a/21805970/18639286 (Though the question was a C++ question, the answer actually quotes C standard specifications)

I have edited the above changes into your code to get the following:

int main (int argc,char* argv[])
{

    char s[100];
    if(scanf("%99s",s)!=1)
    {
       fprintf(stderr, "Invalid or missing input\n");
       // scanf should have returned 1 if it successfully read 1 input
    }
    char *t = malloc(strlen(s) + 1); // to store the letters + '\0' at the end
    if(t==NULL)
    {
       fprintf(stderr, "Memory allocation failed\n");
       exit(1);
    }

    strcpy(t, s);
    t[0] = toupper((unsigned char)t[0]);
    printf("s: %s\n", s);
    printf("t: %s\n", t);

    free(t);
    return 0;

}

Some extra points:

You can also use fgets instead of scanf :
fgets(s, 100, stdin); as fgets does bound checking.

Also, fgets reads till end of line/file or till the buffer is full, so it will read past the spaces in input, But scanf will stop at a space. Or as Ted Lyngmo pointed out in the comments, scanf with %s reads a word, while fgets reads a line.

But, even if you intend to read only a word, you can use fgets and then use sscanf to read the first word from the string with which you used fgets. Example:

char buffer[20];
char str[sizeof buffer]=""; //initialize with an empty string
if(fgets(buffer, sizeof buffer, stdin)==NULL)
{
  fprintf(stderr, "Error reading input\n");
  exit(1);
}
if(sscanf(buffer, "%s", str)!=1)
{
  // if str was not read
  fprintf(stderr, "String not read\n");
  exit(1);
}
// DO stuff with str

Here, you don't need to use %19s because we know the size of buffer.

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