Leetcode问题是"罗马整数"；。这是我的解决方案：
char* intToRoman(int num) {
char numerals[] = { 'M', 'D', 'C', 'L', 'X', 'V', 'I' };
int numeral_vals[] = { 1000, 500, 100, 50, 10, 5, 1 };
int prefix_minus[] = { 100, 100, 10, 10, 1, 1, 0 };
int arr_len = 7;
char* roman = (char*)malloc(sizeof(char) * 100);
int romanLen = 0;
for (int i = 0; i < arr_len; i++) {
while (num >= numeral_vals[i]) {
roman[romanLen] = numerals[i];
romanLen++;
num -= numeral_vals[i];
}
int prefixed_val = numeral_vals[i] - prefix_minus[i];
if (num >= prefixed_val) {
roman[romanLen] = prefixed_val;
romanLen++;
num -= prefixed_val;
}
}
return roman;
}
通过VS代码输入3，它就工作了。但在Leetcode本身，它给了我一个：
=================================================================
==34==ERROR: AddressSanitizer: heap-buffer-overflow on address 0x60b000000154 at pc 0x562764199bbb bp 0x7ffe8128ec40 sp 0x7ffe8128ec30
READ of size 1 at 0x60b000000154 thread T0
#2 0x7f53a881f0b2 in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x270b2)
0x60b000000154 is located 0 bytes to the right of 100-byte region [0x60b0000000f0,0x60b000000154)
allocated by thread T0 here:
#0 0x7f53a9464bc8 in malloc (/lib/x86_64-linux-gnu/libasan.so.5+0x10dbc8)
#3 0x7f53a881f0b2 in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x270b2)
Shadow bytes around the buggy address:
0x0c167fff7fd0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0x0c167fff7fe0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0x0c167fff7ff0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0x0c167fff8000: fa fa fa fa fa fa fa fa fd fd fd fd fd fd fd fd
0x0c167fff8010: fd fd fd fd fd fa fa fa fa fa fa fa fa fa 00 00
=>0x0c167fff8020: 00 00 00 00 00 00 00 00 00 00[04]fa fa fa fa fa
0x0c167fff8030: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0c167fff8040: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0c167fff8050: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0c167fff8060: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0c167fff8070: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
Shadow byte legend (one shadow byte represents 8 application bytes):
Addressable: 00
Partially addressable: 01 02 03 04 05 06 07
Heap left redzone: fa
Freed heap region: fd
Stack left redzone: f1
Stack mid redzone: f2
Stack right redzone: f3
Stack after return: f5
Stack use after scope: f8
Global redzone: f9
Global init order: f6
Poisoned by user: f7
Container overflow: fc
Array cookie: ac
Intra object redzone: bb
ASan internal: fe
Left alloca redzone: ca
Right alloca redzone: cb
Shadow gap: cc
==34==ABORTING
经过一番检查，看起来mallocing很好，但我无法返回"罗马"；。但这怎么了？VS Code处理得很好。
此外，leetcode没有标签，我可以用什么来代替？
在行
roman[romanLen] = prefixed_val;
作为输出字符串的一部分的roman[romanLen]被分配给作为辅助变量的prefixed_val。我不知道这个算法，但这似乎不正确。实际上，对于数字4，函数返回一个包含4的char数组，后面跟99个零。测试(我在函数中将malloc替换为calloc，以确保安全打印整个字符串(：
char *result = intToRoman(4);
for (int k = 0; k < 100; ++k) {
printf("%d,", result[k]);
}
还有其他问题，正如约翰·博林格、迪米奇和退役忍者在对该问题的评论中指出的那样。请记住，malloc不会初始化内存(请参阅man 3 malloc(，但calloc会初始化。因此，字符串roman应该在执行任何您认为它是字符串的操作之前显式地以' '终止。
我不知道Leetcode产生的错误，但也许这些信息无论如何都会对你有所帮助。
顺便说一句，在未来，你可能会考虑使用Valgrind。当涉及到处理内存时，它会产生非常有用的消息。如果使用gcc，请记住将其与标志-g一起使用。此标志使Valgrind能够将行号添加到其输出中。