所以这是我在头文件中的结构:
struct _Variable {
char *variableName;
char *arrayOfElements;
int32_t address;
};
typedef struct _Variable Variable;
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这是我在.c文件中的init函数的实现:
void initVariable(Variable *variable, char *variableName, char *arrayOfElements,
int32_t address) {
int lengthOfVariableNameWithTerminatingChar = strlen(variableName) + 1;
variable->variableName = malloc(
sizeof(char) * lengthOfVariableNameWithTerminatingChar);
strncpy(variable->variableName, variableName,
lengthOfVariableNameWithTerminatingChar);
int lengthOfArrayOfElementsWithTerminatingChar = strlen(arrayOfElements)
+ 1;
variable->arrayOfElements = malloc(
sizeof(char) * lengthOfArrayOfElementsWithTerminatingChar);
strncpy(variable->arrayOfElements, arrayOfElements,
lengthOfArrayOfElementsWithTerminatingChar);
variable->address = address;
}
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我编译时没有错误,但是当我运行我的测试文件时:
void test_initVariable() {
printf("\n---------------test_initVariable()-----------------\n");
// TODO:
Variable *variable1;
initVariable(variable1, "variable1", "1, 2, 3", 4); // <== Causes binary .exe file to not work
}
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谁能告诉我如何修复我的实现?
Variable *variable1;
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给你一个未初始化的指针.你没有它所指向的内存,所以无法安全地写入它.
您需要为其分配存储空间 variable1
Variable variable1;
initVariable(&variable1, "variable1", "1, 2, 3", 4);
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会工作.
如果你想variable1动态分配,最容易initVariable处理它
Variable* initVariable(char *variableName, char *arrayOfElements, int32_t address)
{
Variable* var = malloc(sizeof(*var));
if (var != NULL) {
var->variableName = strdup(variableName);
var->arrayOfElements = strdup(arrayOfElements);
var->address = address;
}
return var;
}
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请注意,我在这里也简化了字符串的分配/填充.您的代码可以工作,但如果您使用的是posix兼容系统,strdup是一种更简单的方法来实现相同的结果.
正如在注释中所讨论的,如果字符串成员Variable都是字符串文字,则不需要分配存储.在这种情况下,您可以简化操作
Variable* initVariable(char *variableName, char *arrayOfElements, int32_t address)
{
Variable* var = malloc(sizeof(*var));
if (var != NULL) {
var->variableName = variableName;
var->arrayOfElements = arrayOfElements;
var->address = address;
}
return var;
}
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