0
我想將C轉換爲x86。我正在使用一個結構...x86結構scanf
struct person_record_struct
{
char last_name[128];
char first_name[128];
char year_of_birth[10];
int month_of_birth; // January => 1
int day_of_birth; // 1st Day of a Month => 1
char drivers_license_no[128];
};
typedef struct person_record_struct person_record;
我無法讓我的scanf工作。這裏是C ..
result = scanf("%s\n%s\n%s\n%d\n%d\n%s\n", &records[counter].last_name[0],
&records[counter].first_name[0], &records[counter].year_of_birth[0],
&records[counter].month_of_birth, &records[counter].day_of_birth,
&records[counter].drivers_license_no[0]);
而且我86 ..
;counter @ [ebp-4]
;records @ [ebp-16]
; format_string_main_2 db '%s\n%s\n%s\n%d\n%d\n%s\n', 0
; read in info
; push drivers_license_no
mov ebx, [ebp-16] ;
mov eax, [ebp-4]
mov ecx, struct_size
mul ecx
add eax, ebx
lea eax, [eax+276]
push eax
; push day_of_birth
mov ebx, [ebp-16]
mov eax, [ebp-4]
mov ecx, struct_size
mul ecx
add eax, ebx
lea eax, [eax+272]
push eax
; push month_of_birth
mov ebx, [ebp-16]
mov eax, [ebp-4]
mov ecx, struct_size
mul ecx
add eax, ebx
lea eax, [eax+268]
push ax
; push year_of_birth
mov ebx, [ebp-16]
mov eax, [ebp-4]
mov ecx, struct_size
mul ecx
add eax, ebx
lea eax, [eax+256]
push eax
; push first_name
mov ebx, [ebp-16]
mov eax, [ebp-4]
mov ecx, struct_size
mul ecx
add eax, ebx
lea eax, [eax+128]
push eax
; push last_name
mov ebx, [ebp-16]
mov eax, [ebp-4]
mov ecx, struct_size
mul ecx
add eax, ebx
lea eax, [eax+0]
push eax
push format_string_main_2
call scanf
add esp, 28
mov [ebp-12], eax
我使用一個檢查，看看結果是6，如果它不是我的程序，打印錯誤並退出。它一直有一個錯誤，我不知道我做錯了什麼。任何幫助將非常感激。謝謝。
這是一個似乎是正確的我的電話釋放calloc ...
; // allocate the buffer of all the records
; records = (person_record *)calloc(number_of_records, sizeof(person_record));
push struct_size
mov eax, [ebp-8]
push eax
call calloc
add esp, 8
mov [ebp-16], eax
學習使用GDB。這對任何彙編程序員來說都是非常有價值的。而且，你有什麼錯誤？ – refi64
注意：在'scanf（）'格式'「...％s \ n」'的末尾，很少有代碼需要空格。建議'「...％s」'。或更好''％127s％127s％9s％d％d％127s「' – chux