小编zz3*_*r90的帖子

简单地说：

1. 栈顶($esp) = 0xbffff49c。
2. gdb 执行ret指令，该指令以Cannot access memory at address 0x90909094.

0x90909094当堆栈顶部的值是 时，gdb 尝试访问的原因是0xbffff49c什么？

随机信息（以防万一）：

[----------------------------------registers-----------------------------------]
EAX: 0x5a ('Z')
EBX: 0xb7fbeff4 --> 0x15ed7c 
ECX: 0xbffff428 --> 0xb7fbf4e0 --> 0xfbad2a84 
EDX: 0xb7fc0360 --> 0x0 
ESI: 0x0
EDI: 0x0 
EBP: 0x90909090 
ESP: 0xbffff49c --> 0xbffff450 --> 0xdb31c031 
EIP: 0x80485dd (<greeting+113>: ret)
EFLAGS: 0x292 (carry parity ADJUST zero SIGN trap INTERRUPT direction overflow)
[-------------------------------------code-------------------------------------]
   0x80485d0 <greeting+100>:    mov    DWORD PTR [esp],0x80487f4
   0x80485d7 <greeting+107>:    call   0x80483f0 …

我正在编写一个toString方法,它打印输出指向结构的指针的所有属性.在阅读处理字符串的安全方法时,我最终创建了以下解决方案:注意Person结构具有属性名称,权重,高度和年龄(除了名称之外的所有整数,这是一个char数组).

char* toString(struct Person* inputPerson)
{
    // Allocate memory to return string
    char* ret = malloc(sizeof (char) * 100);
    // copy "Name: " into string
    strcpy(ret, "Name: ");
    // safely copy name, at most leaving enough room for the other params (leaving 50 bytes) 
    strncat(ret, inputPerson->name, 100-50);
    // copy "Age: " into string
    strcat(ret, "\n\tAge: ");
    // create tmp char to allow us to convert ints age, weight, and height into strings
    char tmp[4];
    // safely convert string to int …