将指向数组的指针传递给另一个函数C
将指向数组的指针传递给另一个函数C
提问于 2015-04-16 03:42:29
我尝试获取一个最大为2048字节的输入文件，并将其放入layer4中自己的数组中。我还尝试将数组的大小放在数组的spot中，以供以后使用。当layer4完成时，我尝试将指向名为code的数组的指针传递给transmit函数，该函数将该值传递给layer3，并将该数组放入一个结构中。目前，当我将layer4和layer3中指针的地址相互比较时，它们是匹配的。但是，当我在layer3中检查数组中的值时，它们与输入文件中数组的值不匹配。这段代码将成为一个更大项目的一部分。我收到的各种警告位于我的代码的底部：
#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
^回答 1
Stack Overflow用户
发布于 2015-04-16 03:58:10
你到处都是指针和整数。这并不总是会造成问题，但充其量也是糟糕的做法。对于您的特定问题，原因可能是：
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;将一个整型指针传递给layer3调用。但是你可以在layer3中获取它的地址。这不是你想要的。同样，int和指针的混合不是这里的根本原因，但确实造成了混乱。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，不要更改指向整数的指针。只需直接传递指针(特别是传递到layer3中)。编译器的警告已经强烈地暗示了这一点。如果您摆脱了所有这些警告，这是您在正确的轨道上的一个好兆头。
更多小贴士：
- 你没有为你的函数声明显式的返回类型。声明正确地explicitly.
- Format你的代码(一致的间距，等等)总是很好的做法。尤其是在Stackoverflow上发帖时，对你自己来说更是如此。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/29659392
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