问题写memset的功能 [英] Problems writing the memset function

问题描述

我写了 memset的函数和我的code是下面，我有一个问题

 无效* memsetFun（void *的指针，INT C中，int的大小）{  如果（指针= NULL＆放大器;！＆安培;尺寸大于0）{    无符号字符* PCHAR =指针;
    INT I = 0;      对于（i = 0; I＆LT;大小; ++ I）{      unsigned char型TEMP =（unsigned char型）C;      * PCHAR ++ =温度; //或PCHAR [I] =温度（他们都没有工作）    }
  }
    返回指针;
}
 

我也试过PCHAR [I] =我们想和仍然没有工作的价值。它给了我一些垃圾数量不作任何意义。

和我打电话吧：

  memsetFun（地址，NUM，大小）;
的printf（％2 P值为％d \\ n，地址，*（（INT *）地址））;
 

在哪里我所说的地址（我只是输入地址）

例如，如果您打印字符（三）打印像一个奇怪的字符，看起来像（为价值4）

  0 0
0 4
 

解决方案

您code看起来很好，我和这里的几个人都评论说，它工作在其系统上。

所以，很明显的事情是调试它 - 这是一个技能，会派上用场颇有几分未来:-)你应该了解它的现在的

什么以下code输出，当你运行它？

 无效* memsetFun（void *的指针，INT C中，int的大小）{
    的printf（A％X％D \\ n，指针，C大小）;
    如果（指针= NULL＆放大器;！＆安培;尺寸大于0）{
        的printf（B \\ n）;
        无符号字符* PCHAR =指针;
        INT I = 0;
        对于（i = 0; I＆LT;大小; ++ I）{
            的printf（C％D（％D），我，* PCHAR）;
            unsigned char型TEMP =（unsigned char型）C;
            * PCHAR ++ =温度; //或PCHAR [I] =温度（他们都没有工作）
            的printf（ - ＆GT;（％D），我*（PCHAR-1））;
        }
    }
    的printf（D \\ n）;
    返回指针;
}
 

从输出中，应该明确的code正在采取什么样的路径，你的参数是什么（这将大大有助于调试过程）。

的更新：的

如果你与1比0以外的任何尽显你的内存块，并使用这样的：

 的printf（％2 P值为％d \\ n，地址，*（（INT *）地址））;
 

要打印出来，你的将会的得到奇怪的结果。

你基本上询问了一些这些字节是PTED为整数间$ P $。因此，举例来说，如果你用 0×02 字节填充它，你有一个4字节的整数类型，你会得到整数 0x02020202 （33686018）的不的 0×02 ，你可能期望。如果你想看看第一的字符的值，使用：

 的printf（％2 P值为％d \\ n，地址，*（（字符*）地址））;
 

和基于您的最新的的问题更新：

  

例如，如果您打印字符（三）打印像一个奇怪的字符，看起来像（该值4）结果
  0 0结果
  0 4

如果这是一个单一的性格和你打印它作为一个字符，则可能是没有错的。许多输出流会给你，对于一个控制字符（CTRL-D在这种情况下，ASCII code 4）。如果你不是用ASCII code的0x30（48），你会看到的字符'0'或ASCII×41（65）填充它会给你'A'。

I am writing the memset function and my code is below, I am having a problem

void* memsetFun(void* pointer, int c, int size) {

  if ( pointer != NULL && size > 0 ) {

    unsigned char* pChar =  pointer;


    int i = 0;

      for ( i = 0; i < size; ++i) {

      unsigned char temp = (unsigned char) c;

      *pChar++ = temp; // or pChar[i] = temp (they both don't work)

    }
  }  
    return pointer;


}

I also tried pChar[i] = the value we want and still not working. It gives me some trash numbers that do not make any sense.

And I am calling it:

memsetFun(address, num, size);
printf("value at %p is %d\n", address, *((int*) address));

Where I call the address (I just input the address)

For example, if you to print the chars ( c ) it prints like a weird char that looks like ( for the value 4 )

0 0
0 4

解决方案

Your code looks fine to me and several people here have commented that it works on their system.

So the obvious thing to do is to debug it - that's a skill that will come in handy quite a bit in future :-) You should learn it now.

What does the following code output when you run it?

void* memsetFun(void* pointer, int c, int size) {
    printf("A %x %d %d\n", pointer, c, size);
    if ( pointer != NULL && size > 0 ) {
        printf("B\n");
        unsigned char* pChar =  pointer;
        int i = 0;
        for ( i = 0; i < size; ++i) {
            printf("C %d (%d)", i, *pChar);
            unsigned char temp = (unsigned char) c;
            *pChar++ = temp; // or pChar[i] = temp (they both don't work)
            printf(" -> (%d)", i, *(pChar-1));
        }
    }  
    printf("D\n");
    return pointer;
}

From the output, it should be clear what paths the code is taking and what your parameters are (which will greatly assist the debugging process).

Update:

If you're filling your memory block with anything other than zeros and using this:

printf("value at %p is %d\n", address, *((int*) address));

to print it out, you will get strange results.

You're basically asking for a number of those bytes to be interpreted as an integer. So, for example, if you filled it with 0x02 bytes and you have a 4-byte integer type, you will get the integer 0x02020202 (33686018), not 0x02 as you may expect. If you want to see what the first character value is, use:

printf("value at %p is %d\n", address, *((char*) address));

And based on your latest question update:

For example, if you to print the chars ( c ) it prints like a weird char that looks like ( for the value 4 )
0 0
0 4

If that's a single character and you're printing it as a character, there's probably nothing wrong at all. Many output streams will give you that for a control character (CTRL-D in this case, ASCII code 4). If you instead filled it with ASCII code 0x30 (48), you would see the character '0' or ASCII 0x41 (65) would give you 'A'.

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