Return 函数中的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"
}

函数 printTest 似乎没有打印出任何东西，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，该函数在初始化值后确实有效。

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

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

Answer 1

您没有从测试中获取 return 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;
}

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，您将获取分配的指针，并使用 printf 将格式化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;
}

Answer 2

此外，sprintf 输出到一个字符串。它不输出到标准输出，即打印到屏幕，不像 printf 那样。您可能想在 output 字符串上调用 printf 或 puts。

Answer 3

问题比乍一看要多得多。这是一个清理过的版本。您不能分配字符串（例如 mem->name = name;）。您可以为 mem->name 分配，然后为它分配 strncpy 名称，或者使用 strdup 分配并立即复制。在 printTest 中，您必须为 output 分配 space 足以容纳格式字符串的静态部分 plus print->name 和 print->num。请查看以下内容，如果您有任何问题，请告诉我。

#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"
}

输出

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

Answer 4

这一行：'sprintf(output,"It's name is %s, and it costs %d",print->name,print->num)' 需要进行一些重大修改。 1) 源字段" print->name 和 print->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' 指向一个分配的内存区域，但是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文件的大小结构，sprintf 格式参数中的所有额外字符将没有空间。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

Return 函数中的语句从不打印到屏幕

Return statement in function never prints to screen

c

memory

struct

pointers