串联在C字符数组：动态内存处理 [英] Concatenating a char array in C: dynamically handling memory
问题描述
我对我在做什么错在我的C程序困惑：我试图创建一个 开始，增加了6串'！值从传感器（用逗号隔开）读取，然后发送它的串行端口。示例输出将是：！5,5,5,5,5,5或34,34,34,34，！ 34,34。
问题：由于传感器的值（在上面的例子中5或34）的范围从0到255，我也不在运行时知道我的字符数组需要有多大是。这意味着我要我要添加到我的字符串每次动态重新分配内存。下面是我尝试这样做，但我这样做是错误的，因为我什么也看不到在我的串行端口来（表示有一个运行时错误）。
我怎样才能正确地实现code为字符串动态分配内存？我试图使用的malloc 和的realloc 的行为不符合市场预期。
的char * convertIntToString（uint8_t有integerValue）{
字符*海峡=的malloc（4）; //最多3位数+ 1为空终止
utoa（integerValue，STR，10）;
返回海峡;
}字符* CONCAT（字符* S1，字符* S2）
{
字符*结果=的malloc（strlen的（S1）+ strlen的（S2）+1）; // + 1零终止
//在实际code，你会在malloc的检查错误在这里
的strcpy（结果是，S1）;
的strcat（结果，S2）;
返回结果;
}INT主要（无效）
{
uint8_t有analogValue;
字符* outputStr =的malloc（1）; //字符数组= 1元initalize大小 而（1）{
outputStr = realloc的（outputStr，1）;
outputStr = CONCAT（outputStr，！）;
analogValue = ReadADC（0）;
outputStr = CONCAT（outputStr，convertIntToString（analogValue））;
的for（int i = 0;我小于5;我++）{
outputStr =的concat（outputStr，，）;
outputStr = CONCAT（outputStr，convertIntToString（analogValue））;
}
CDC_Device_SendString（安培; VirtualSerial_CDC_Interface，outputStr）; //通过USB发送字符串
免费（outputStr）;
}
}
您正在运行到不确定的行为，因为 outputStr 的内容不会在第一次声明中正确初始化内在，而循环。
outputStr = realloc的（outputStr，1）; // outputStr未初始化。
他们更改为：
outputStr = realloc的（outputStr，2）;
的strcpy（outputStr，！）;
您也渗出了一大堆的内存。从 convertToString 返回的值是永远免费 D。
您可以通过改变策略，一点点避免这样的问题。
改变功能到期望的字符串并使用它。
的char * convertIntToString（uint8_t有integerValue，
字符*海峡）
{
utoa（integerValue，STR，10）;
返回海峡;
}
然后，更改其用法为：
outputStr = CONCAT（outputStr，convertIntToString（analogValue，STR））;
您也泄漏内存，由于使用的是 CONCAT 。
outputStr = CONCAT（outputStr，）;
这泄漏 outputStr 的旧值。你需要保持 outputStr 的旧价值的时间长一点，所以你可以免费了。
下面是我为，而环路建议：
而（1）{ outputStr = realloc的（outputStr，2）;
的strcpy（outputStr，！）; analogValue = ReadADC（0）; 炭海峡[4]; //这是你所需要的最大。
//没有必要malloc和自由。 outputStr = CONCAT（outputStr，convertIntToString（analogValue，STR））; 的for（int i = 0;我小于5;我++）{ 字符* =中newstr CONCAT（outputStr，）; //使用新的内存之前释放旧内存
免费（outputStr）;
outputStr =中newstr; =中newstr CONCAT（outputStr，convertIntToString（analogValue，STR））; //使用新的内存之前释放旧内存
免费（outputStr）;
outputStr =中newstr;
}
CDC_Device_SendString（安培; VirtualSerial_CDC_Interface，outputStr）; //通过USB发送字符串
免费（outputStr）;
}
I'm confused on what I'm doing wrong in my C program: I'm trying to create a string that begins with a '!' and adds 6 values read from a sensor (separated by commas) and then sends it over a serial port. A sample output would be: "!5,5,5,5,5,5" or "!34,34,34,34,34,34".
The problem: Because the sensor values (5 or 34 in the example above) can range from 0 to 255, I don't know at runtime how big my char array needs to be. This means I have to dynamically reallocate memory every time I want to add to my string. Below is my attempt to do so, but I'm doing this wrong because I see nothing coming across my serial port (indicating that there's a runtime error).
How can I properly implement code to allocate memory dynamically for a string? My attempts to use malloc and realloc aren't behaving as expected.
char* convertIntToString(uint8_t integerValue){
char *str = malloc(4); //up to 3 digits + 1 for null termination
utoa(integerValue, str, 10);
return str;
}
char* concat(char *s1, char *s2)
{
char *result = malloc(strlen(s1)+strlen(s2)+1);//+1 for the zero-terminator
//in real code you would check for errors in malloc here
strcpy(result, s1);
strcat(result, s2);
return result;
}
int main(void)
{
uint8_t analogValue;
char *outputStr = malloc(1); //initalize size of char array = 1 element
while (1) {
outputStr = realloc(outputStr, 1);
outputStr = concat(outputStr, "!");
analogValue = ReadADC(0);
outputStr = concat(outputStr, convertIntToString(analogValue));
for(int i = 0; i < 5; i++){
outputStr = concat(outputStr, ",");
outputStr = concat(outputStr, convertIntToString(analogValue));
}
CDC_Device_SendString(&VirtualSerial_CDC_Interface, outputStr); //send string via USB
free(outputStr);
}
}
You are running into undefined behavior since the contents of outputStr are not initialized properly in the first statement inside the while loop.
outputStr = realloc(outputStr, 1); // outputStr is not initialized.
Change them to:
outputStr = realloc(outputStr, 2);
strcpy(outputStr, "!");
You are also leaking a whole bunch of memory. The value returned from convertToString is never freed.
You can avoid that problem by changing the strategy a little bit.
Change the function to expect a string and use it.
char* convertIntToString(uint8_t integerValue,
char* str)
{
utoa(integerValue, str, 10);
return str;
}
Then, change its usage as:
outputStr = concat(outputStr, convertIntToString(analogValue, str));
You are also leaking memory due to the way you are using concat.
outputStr = concat(outputStr, ",");
That leaks the old value of outputStr. You need to keep the old value of outputStr for a bit longer so you can free it.
Here's my suggestion for the while loop:
while (1) {
outputStr = realloc(outputStr, 2);
strcpy(outputStr, "!");
analogValue = ReadADC(0);
char str[4]; // This is the max you need.
// There is no need to malloc and free.
outputStr = concat(outputStr, convertIntToString(analogValue, str));
for(int i = 0; i < 5; i++){
char* newStr = concat(outputStr, ",");
// free the old memory before using the new memory
free(outputStr);
outputStr = newStr;
newStr = concat(outputStr, convertIntToString(analogValue, str));
// free the old memory before using the new memory
free(outputStr);
outputStr = newStr;
}
CDC_Device_SendString(&VirtualSerial_CDC_Interface, outputStr); //send string via USB
free(outputStr);
}
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