c - 将指向数组的指针传递给另一个函数 C

Question

我正在尝试获取一个最多 2048 字节的输入文件并将其放置在第 4 层中它自己的数组中。我还试图将数组的大小放在数组的 spot [0] 中以供以后使用。当 layer4 完成时，我试图将指向名为 code 的数组的指针传递给 transmit 函数，它将将该值传递给 layer3 并放置结构中的数组。目前，当我将 layer4 和 layer3 中的指针地址相互比较时，它们是匹配的。但是，当我检查 layer3 中数组中的值时，它们与输入文件中数组的值不匹配。此代码将成为更大项目的一部分。我收到的各种警告位于我的代码下方:

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


main()
{
int *senddata;
senddata = layer4(); // get pointer address of input array
transmit(senddata); //put pointer value into transmit


}

int layer4(){
    FILE *file = fopen("sendtext.txt", "r"); //
    char *code;
    size_t n = 0;
    int c;
    if (file == NULL)
        return NULL; //could not open file

    code = malloc(2048); //allocate memory 

    while ((c = fgetc(file)) != EOF)
    {
    n++;       
    code[n] = (char) c;
    printf("%c", code[n]);
    }
    code[n] = '\0';
    n = n-1;  // for some reason the byte size is +1 for what it should be
    code[0] = n;  

  printf("Check Pointer Address in layer 4: %p \n", code); //test to see pointer address
  printf("Check to see value in pointer:%c \n", code[0]); //check to see if the byte size was placed in the array
  printf("Byte size:%zd\n", n); /// see array size


return code;
}

transmit(int* getdata){  //gets pointer value
int newdata = getdata;
int g = layer3(newdata); //puts pointer into new function
}

layer3(int b){
int x = b;
int w = &x;
char *MSS;
MSS = malloc(60); 


printf("Check to see value in pointer:%c \n", w);


printf("Check Pointer Address in layer 3:%p \n", x); //test to see pointer address

struct l3hdr {  
  char ver;
  char src_ipaddr[16];
  char dest_ipaddr[16];
  char reserved[7];
};

struct l3pdu {
// put array here
struct l3hdr hdr3; 


};
}

输出

Q sadfasd fsa asd fsadf sad f /// This is my input testfile
Check Pointer Address in layer 4: 0xa81250 
Check to see Byte size in array: 
Check to see first input character in array:Q 
Byte size:29
Check to see value in pointer:P 
Check Pointer Address in layer 3:0xa81250   

警告

lab.c:20:9: warning: return makes integer from pointer without a cast [enabled by default]
         return NULL; //could not open file
         ^
lab.c:37:2: warning: format ‘%c’ expects argument of type ‘int’, but argument 2 has type ‘char *’ [-Wformat=]
  printf("Check to see value in pointer:%c \n", code);
  ^
lab.c:43:1: warning: return makes integer from pointer without a cast [enabled by default]
 return code;
 ^


lab.c: In function ‘transmit’:
lab.c:47:15: warning: initialization makes integer from pointer without a cast [enabled by default]
 int newdata = getdata;
               ^


lab.c: In function ‘layer3’:



lab.c:64:9: warning: initialization makes integer from pointer without a cast [enabled by default]
 int w = &x;
         ^


lab.c:72:1: warning: format ‘%p’ expects argument of type ‘void *’, but argument 2 has type ‘int’ [-Wformat=]
 printf("Check Pointer Address in layer 3:%p \n", x); //test to see pointer address
 ^

Answer 1

你到处都是指针和整数。这并不总是会导致问题，但充其量是一种不好的做法。对于您的具体问题，原因可能是这样的:

transmit(int* getdata){  //gets pointer value
    int newdata = getdata;
    int g = layer3(newdata); //puts pointer into new function
}

layer3(int b){
    int x = b;
    int w = &x;

您将一个 int 指针传递给 layer3 调用。但是你把它的地址放在 layer3 里面。那不是你想要的。同样，混合使用 int 和指针并不是这里的根本原因，但确实造成了困惑。你的代码应该是这样的:

transmit(int* getdata){  //gets pointer value
    int g = layer3(getdata); //puts pointer into new function
}

layer3(int *b){
    int *w = b;

也就是说，不要将指针更改为整数。只需将指针直接传递(特别是传递到第 3 层)。编译器警告已经向您强烈暗示了这一点。如果您摆脱了所有这些警告，那么表明您走在正确轨道上的一个好兆头。

更多提示:

1. 您没有为您的函数声明明确的返回类型。始终明确声明这一点的好习惯。
2. 正确格式化您的代码(一致的间距等)。尤其是在 Stackoverflow 上发帖时，但对您自己来说更是如此。