Question

我试图将一些参数传递到一个函数中，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。

这是我想要做的事情：

#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"
}

函数printTest，似乎没有打印出任何内容，相反，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，这在初始化值后起作用。

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

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

Answer 1

您没有捕获从测试中返回的值，因此它会丢失：

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

您的打印功能也是错误的：

// 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，你将获取你分配的指针，并使用printf将格式化的字符串写入stdout：

// 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;
}

还有关于测试功能的一些注意事项：

// 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;
}

请改为尝试：

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;
}

Answer 2

乍一看还有很多问题。这是一个清理版本。您无法指定字符串（例如mem->name = name;）。您可以为其分配mem->name然后strncpy名称，也可以使用strdup一次分配和复制。在printTest中，您必须为output分配空间，足以容纳格式字符串加上 print->name和print->num的静态部分。如果您有疑问，请查看以下内容并告诉我。

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

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

temp * test(char * name, int num);
char * printTest(temp * print);

int main (void)
{
    struct temp * t = test("TV",200);
    // struct temp * t;
    printf ("%s\n", 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->name = strdup (name);

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

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

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

    output = malloc (strlen (print->name) + sizeof print->num + 30);

    sprintf(output,"It's name is %s, and it costs %d",print->name,print->num);

    return output; //should print "TV" and costs "200"
}

<强>输出

$ ./bin/struct_rtn
It's name is TV, and it costs 200

Answer 3

此外，sprintf输出到字符串。它不输出到标准输出，即打印到屏幕，不像printf那样。您可能希望在printf字符串上拨打puts或output。

Answer 4

这一行：'sprintf（输出，“它的名称是％s，它的成本％d”，print-＆gt; name，print-＆gt; num）'需要一些重要的修改。 1）源字段“print-＆gt; name和print-＆gt; num在分配的内存段中但是，它们从未被设置为任何特定值，因此它们包含垃圾。目标指针'output'指向分配内存区域，但该区域不大于源区域（通常，它应该被格式化为'temp'结构，而不是一个长字符串。并且因为它只是临时结构的大小， sprintf格式参数中的所有额外字符都没有空间。结果将是未定义的行为，因为sprintf的输出将超出分配的内存区域

函数中的return语句永远不会打印到屏幕上

4 个答案: