c - 函数中的返回语句从不打印到屏幕

Question

我试图将一些参数传递给一个函数，将这些值存储到结构中的一组元素中。然后通过调用另一个函数从结构中打印这些值。

这是我正在尝试做的事情:

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

typedef struct temp
{
    int num;
    char * name;
    struct temp * nextPtr;
}temp;

temp * test();
char * printTest(temp * print);

int main (void)
{
    test("TV",200);
    struct temp * t;
    printTest(t);
    return 0;
}

temp * test(char * name, int num)
{
    struct temp * mem = malloc(sizeof(struct temp));
    mem->name = malloc(sizeof(strlen(name) + 1));
    mem->name = name;

    mem->num = num;
    mem->nextPtr = NULL;
    return mem;
}

char * printTest(temp * print)
{
    char * output;

    output = malloc(sizeof(struct temp));
    print = malloc(sizeof(struct temp));

    sprintf(output,"It's name is %s, and it costs %d",print->name,print->num);
    return output; //should print "TV" and costs "200"
}

printTest 函数似乎没有打印出任何内容，如果我在其中硬核 printf，它会打印空值和零。但是，我尝试在 test 函数中打印结构值，该函数在初始化值后确实有效。

例如，如果我执行 printf("%d",mem->num); 这会打印出 200 的值(在 test 函数中) .

但是最后一个函数中的 sprintf 和 return 组合不会产生相同的结果。任何帮助将不胜感激!

Answer 1

您没有捕获从测试返回的值，因此它只是丢失了:

int main (void)
{
    //changed to capture return value of test.
    struct temp * t = test("TV",200);
    printTest(t);
    return 0;
}

还有你的打印功能是错误的:

// this now points to the struct you allocated in test.
char * printTest(temp * print)
{
    char * output;

    // you don't really want the size of temp here, you want the size
    // of your formatted string with enough room for all members of the 
    // struct pointed to by temp.
    output = malloc(sizeof(struct temp));

    // you're not using this.
    print = malloc(sizeof(struct temp));

    // you sprintf from a buffer pointing to nothing, into your output buffer
    // writing past the memory you actually allocated.
    sprintf(output,"It's name is %s, and it costs %d",print->name,print->num);

    // return doesn't print anything, it simply returns the value that your
    // function signature specifies, in this case char *
    return output; //should print "TV" and costs "200"
}

试试这个，您将获取您分配的指针，并使用 printf 将格式化的字符串写入标准输出:

// we're not returning anything
void printTest(temp * print ){
    if (temp == NULL ){
        // nothing to print, just leave.
        return;
    }

    printf("It's name is %s, and it costs %d",print->name,print->num);
    return;
}

还有一些关于你的测试函数的注释:

// char is a pointer to a string, from your invocation in main, this is
// a string stored in static read only memory
temp * test(char * name, int num)
{
    // you malloc memory for your struct, all good.
    struct temp * mem = malloc(sizeof(struct temp));

    // you malloc space for the length of the string you want to store.
    mem->name = malloc(sizeof(strlen(name) + 1));
    // here's a bit of an issue, mem->name is a pointer, and name is a pointer.
    // what you're doing here is assigning the pointer name to the variable 
    // mem->name, but you're NOT actually copying the string - since you 
    // invoke this method with a static string, nothing will happen and
    // to the string passed in, and you'll be able to reference it - but
    // you just leaked memory that you allocated for mem->name above.
    mem->name = name;

    // num is not apointer, it's just a value, therefore it's okay to assign
    // like this.
    mem->num = num;
    mem->nextPtr = NULL;
    return mem;
}

试试这个:

temp * test(char * name, int num)
{
    struct temp * mem = malloc(sizeof(struct temp));

    // we still malloc room for name, storing the pointer returned by malloc
    // in mem->name
    mem->name = malloc(sizeof(strlen(name) + 1));
    // Now, however, we're going to *copy* the string stored in the memory location
    // pointed to by char * name, into the memory location we just allocated, 
    // pointed to by mem->name
    strcpy(mem->name, name);

    mem->num = num;
    mem->nextPtr = NULL;
    return mem;
}