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首页 > 解决方案 > 为什么我不能返回 realloc 的结果？（当你看到代码时你会得到问题......）

c - 为什么我不能返回 realloc 的结果？（当你看到代码时你会得到问题......）

问题描述

任务是制作 2 个不同的功能。1 其中 malloc 为 char 创建一个长度并放入它们的 2 个字符串。另一个函数不能给出返回类型。它应该改变 2 个融合字符串的长度以适应第三个。但在某些情况下，它并没有给出应有的结果。第一个“失败”应该是一个，但其他 3 个一次不应该是一个......

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

char *concatStrings( char* concaStr,char*str);
void concatStrings2( char* newString,char*str);
void testConcatStrings(char* concaStr,char*str, char*str2, char*expected);

int main()
{
    char *str="Dynamic";
    testConcatStrings("Memory", str,"Management","DynamicMemoryManagement");
    testConcatStrings("Memory ", str,"Management","DynamicMemoryManagement");
    testConcatStrings("Memory", str,"","DynamicMemory");
    testConcatStrings("Memory","" ,"Management","MemoryManagement");
    testConcatStrings("", str,"Management","DynamicManagement");
    testConcatStrings("Memory", "","","Memory");
    testConcatStrings("", str,"",str);
    testConcatStrings("", "","Management","Management");
    testConcatStrings("", "","","");

    return 0;
}

char *concatStrings( char* concaStr,char*str){
    //calculate suze for memory...
    int lenStr=strlen(str);
    int lenConcaStr=strlen(concaStr);
    //+1 byte for 0x00
    int len=lenStr+lenConcaStr+1;
    //reserve memory
    char *result =malloc(len);
    //concatenate string
    memcpy( result,str,lenStr);
    memcpy( result+lenStr,concaStr,lenConcaStr+1);
    printf("Hoi");
    return result;
}

void concatStrings2( char* newString,char*str){
    //calculate suze for memory...
    int lenStr=strlen(str);
    int lenConcaStr=strlen(newString);
    //+1 byte for 0x00
    int len=lenStr+lenConcaStr+1;
    //reserve memory
    printf("\nLen: %d",len);
    newString=(char*)realloc(newString,len);
    printf(" %s %s ",newString,str);
    //concatenate string
    memcpy(newString+lenConcaStr,str,lenStr+1);
    printf(" %s %s ",newString,str);
}

void testConcatStrings(char* concaStr,char*str, char*str2, char*expected){
    //concatenated string by dynamic memory management(alloc)
    char *result = concatStrings(concaStr,str);
    concatStrings2(result,str2);
    printf("\nstr= %s, concat: %s, str2: %s--> result:%s -->", str,concaStr,str2,result);
    //check result
    (strcmp(result,expected)==0)?printf("Success"): printf("Failure");
    //never forget free!!!!
    free(result);
    printf("\nIs free\n");
}

结果：

Hoi
Len: 24 DynamicMemory Management  DynamicMemoryManagement Management
str= Dynamic, concat: Memory, str2: Management--> result:DynamicMemoryManagement -->Success
Is free
Hoi
Len: 25 DynamicMemory  Management  DynamicMemory Management Management
str= Dynamic, concat: Memory , str2: Management--> result:DynamicMemory Management -->Failure
Is free
Hoi
Len: 14 DynamicMemory   DynamicMemory
str= Dynamic, concat: Memory, str2: --> result:DynamicMemory -->Success
Is free
Hoi
Len: 17 Memory Management  MemoryManagement Management
str= , concat: Memory, str2: Management--> result:╚d -->Failure
Is free
Hoi
Len: 18 Dynamic Management  DynamicManagement Management
str= Dynamic, concat: , str2: Management--> result: -->Failure
Is free
Hoi
Len: 7 Memory   Memory
str= , concat: Memory, str2: --> result:Memory -->Success
Is free
Hoi
Len: 8 Dynamic   Dynamic
str= Dynamic, concat: , str2: --> result:Dynamic -->Success
Is free
Hoi
Len: 11  Management  Management Management
str= , concat: , str2: Management--> result:╚d -->Failure
Is free
Hoi
Len: 1
str= , concat: , str2: --> result: -->Success
Is free

（结果看起来很混乱，因为我试图确切地查看函数中发生的情况以及错误的确切开始位置。）

标签： cstringpointersdynamic-memory-allocationrealloc

解决方案

在concatStrings2变量newString中是局部的。如果 realloc 改变了堆上的位置，那么新的位置不会传回给被调用者；您需要明确返回新位置。您之所以成功，是因为 realloc 碰巧能够将 char 数组保持在原位（并放大它）。尝试newString在 realloc 之前和之后打印地址，您会发现当 realloc 更改位置时您会发生故障。

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