[英]Printf - access violation reading location - C++
0xC0000005:訪問沖突讀取位置0xcccccccc。
printf拋出此異常。
我不知道為什么會發生這種情況......這些字符串變量中有值。 我使用printf錯了嗎?
救命! (請看開關盒)
string header;
string body;
string key;
if (!contactList.isEmpty()) {
cout << "Enter contact's name: ";
getline(cin, key);
Contact * tempContact = contactList.get(key);
if (tempContact != NULL) {
string name = tempContact->getName();
string number = tempContact->getNumber();
string email = tempContact->getEmail();
string address = tempContact->getAddress();
//I've just put this here just to test if the variables are being initialized
cout << name + " " + number + " " + email + " " + address << endl;
switch (type) {
case 1:
printf("%-15s %-10s %-15s %-15s\n", "Name", "Number", "Email", "Address");
printf("%-15s %-10s %-15s %-15s\n", name, number, email, address);
break;
case 2:
printf("%-15s %-10s\n", "Name", "Number");
printf("%-15s %-10s\n", name, number);
break;
case 3:
printf("%-15s %-15s\n", "Name", "Email");
printf("%-15s %-15s\n", name, email);
break;
case 4:
printf("%-15s %-15s\n", "Name", "Address");
printf("%-15s %-15s\n", name, address);
break;
default:
printf("%-15s %-10s %-15s %-15s\n", "Name", "Number", "Email", "Address");
printf("%-15s %-10s %-15s %-15s\n", name, number, email, address);
}
} else {
cout << "\"" + key + "\" not found.\n" << endl;
wait();
}
} else {
cout << "Contact list is empty.\n" << endl;
wait();
}
第一個printf打印正常,但第二個將拋出異常,看起來無論我如何傳遞字符串值。
printf的“%s”期望char*
作為參數,而不是std::string
。 因此printf會將您的字符串對象解釋為指針,並嘗試訪問該對象的第一個sizeof(char*)
字節給出的內存位置,這會導致訪問沖突,因為這些字節實際上不是指針。
要么使用字符串的c_str
方法來獲取char*
s,要么不使用printf。
C ++ string
不是printf
對%s
說明符所期望%s
- 它想要一個空終止的字符數組。
您需要使用iostream
作為輸出( cout << ...
)或將字符串轉換為字符數組,例如c_str()
。
sepp2k給出了正確的答案,但我要補充一點:如果你打開完全警告(推薦),編譯器會警告你:
a.cc:8: warning: format ‘%s’ expects type ‘char*’, but argument 2 has type ‘int’
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