正如标题所讲述的故事。下面粘贴的是这段代码和所有依赖项。这个错误对我来说很奇怪,因为我能够在“监视”窗口中访问相同的变量,并且我找不到在运行时出现此错误的有效原因。我能找到的大多数具有相同标题的问题都与访问空地址有关,但这里的情况并非如此。 0X00000069 是无法访问的位置。我还在下面附上了“观看”窗口的屏幕截图。
typedef struct
{
void * variable; // variable address
char * signature;
char name[30];
char type[50];
} GlobalVariable;
GlobalVariable VariableTable[] =
{
{ &a, "int a = 7;", "a", "int"},
{ &ch, "char ch = 'a';", "ch", "char"},
{ NULL, NULL }
};
void PrintGlobalVariables()
{
GlobalVariable * variable = VariableTable;
int count = 0;
while(variable->variable)
{
count++;
variable++;
}
if(!count)
{
OutputDebugString("No global variables passed.\n");
return;
}
int i = 0 ;
variable = VariableTable;
char temp[MAX_PATH];
strcpy(temp, " ");
OutputDebugString("Global Variables = ");
while(variable->variable)
{
if(strcmp(trimwhitespace(variable->type),"int") == 0)
{
////////THIS IS THE STATEMENT BELOW WHERE I AM GETTING ERROR
wsprintf(temp, "%s=%ld",*(char *)variable->signature, *(long *)variable->variable);
} else
if(strcmp(trimwhitespace(variable->type),"char") == 0)
{
wsprintf(temp, "%s=%c", *(char *)variable->name, *(char *)variable->variable);
}
if((variable+1)->variable != NULL)
{
wsprintf(temp + strlen(temp), ",", "");
}
OutputDebugString(temp);
variable++;
}
OutputDebugString("\n");
}
char *trimwhitespace(char *str)
{
char *end;
// Trim leading space
while(isspace(*str)) str++;
if(*str == 0) // All spaces?
return str;
// Trim trailing space
end = str + strlen(str) - 1;
while(end > str && isspace(*end)) end--;
// Write new null terminator
*(end+1) = '\0';
return str;
}
最佳答案
wsprintf(temp, "%s=%ld",*(char *)variable->signature, *(long *)variable->variable);
signature
字段是一个 char *
。取消引用它会给出一个char
。但 %s
格式说明符需要 char *
。解决方法是通过删除取消引用来更改该行(并且强制转换也是不必要的)。
wsprintf(temp, "%s=%ld", variable->signature, *(long *)variable->variable);
关于c - 访问冲突读取位置 0x000000xx - c 结构属性 - Visual Studio,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/37933466/