Remove old interface for mem_move_to_left
Signed-off-by: Dave Rodgman <dave.rodgman@arm.com>
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debf8679e0
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2 changed files with 1 additions and 21 deletions
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@ -327,7 +327,7 @@ void mbedtls_ct_mpi_uint_cond_assign(size_t n,
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#if defined(MBEDTLS_PKCS1_V15) && defined(MBEDTLS_RSA_C) && !defined(MBEDTLS_RSA_ALT)
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void mbedtls_ct_mem_move_to_left(void *start,
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void mbedtls_ct_memmove_left(void *start,
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size_t total,
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size_t offset)
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{
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@ -236,26 +236,6 @@ void mbedtls_ct_memcpy_offset(unsigned char *dest,
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*/
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unsigned mbedtls_ct_size_gt(size_t x, size_t y);
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/** Shift some data towards the left inside a buffer.
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*
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* `mbedtls_ct_mem_move_to_left(start, total, offset)` is functionally
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* equivalent to
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* ```
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* memmove(start, start + offset, total - offset);
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* memset(start + offset, 0, total - offset);
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* ```
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* but it strives to use a memory access pattern (and thus total timing)
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* that does not depend on \p offset. This timing independence comes at
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* the expense of performance.
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*
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* \param start Pointer to the start of the buffer.
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* \param total Total size of the buffer.
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* \param offset Offset from which to copy \p total - \p offset bytes.
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*/
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void mbedtls_ct_mem_move_to_left(void *start,
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size_t total,
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size_t offset);
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#endif /* defined(MBEDTLS_PKCS1_V15) && defined(MBEDTLS_RSA_C) && !defined(MBEDTLS_RSA_ALT) */
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